title stringlengths 1 100 | titleSlug stringlengths 3 77 | Java int64 0 1 | Python3 int64 1 1 | content stringlengths 28 44.4k | voteCount int64 0 3.67k | question_content stringlengths 65 5k | question_hints stringclasses 970
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PYTHON SIMPLE DP ᓚᘏᗢ | form-largest-integer-with-digits-that-add-up-to-target | 0 | 1 | # Approach\ndp which is easy to see and read :) <3\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n\n# Code\n```\nclass Solution:\n def largestNumber(self, cost: List[int], target: int) -> str:\n dp=[""]*(target+1)\n for i in range(0,target+1):\n for j in range(1,10... | 0 | Given two positive integers `n` and `k`, the binary string `Sn` is formed as follows:
* `S1 = "0 "`
* `Si = Si - 1 + "1 " + reverse(invert(Si - 1))` for `i > 1`
Where `+` denotes the concatenation operation, `reverse(x)` returns the reversed string `x`, and `invert(x)` inverts all the bits in `x` (`0` changes to ... | Use dynamic programming to find the maximum digits to paint given a total cost. Build the largest number possible using this DP table. |
Python DP Solution | Easy to Understand | Faster than 75% | form-largest-integer-with-digits-that-add-up-to-target | 0 | 1 | # Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe follow a simple take/leave type Memoization DFS where we choose either to take or leave a certain digit. If it\'s not possibe, we return a generic string ```"N"```. Then we compare strings formed in both the cases and return the largest... | 0 | Given an array of integers `cost` and an integer `target`, return _the **maximum** integer you can paint under the following rules_:
* The cost of painting a digit `(i + 1)` is given by `cost[i]` (**0-indexed**).
* The total cost used must be equal to `target`.
* The integer does not have `0` digits.
Since the ... | Use the maximum length of words to determine the length of the returned answer. However, don't forget to remove trailing spaces. |
Python DP Solution | Easy to Understand | Faster than 75% | form-largest-integer-with-digits-that-add-up-to-target | 0 | 1 | # Approach\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe follow a simple take/leave type Memoization DFS where we choose either to take or leave a certain digit. If it\'s not possibe, we return a generic string ```"N"```. Then we compare strings formed in both the cases and return the largest... | 0 | Given two positive integers `n` and `k`, the binary string `Sn` is formed as follows:
* `S1 = "0 "`
* `Si = Si - 1 + "1 " + reverse(invert(Si - 1))` for `i > 1`
Where `+` denotes the concatenation operation, `reverse(x)` returns the reversed string `x`, and `invert(x)` inverts all the bits in `x` (`0` changes to ... | Use dynamic programming to find the maximum digits to paint given a total cost. Build the largest number possible using this DP table. |
Easy python solution 🎉 | number-of-students-doing-homework-at-a-given-time | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->First we run a loop till the length of any one the given array.Then as we know the length of both the arrays are equal so we check the elements of both array at once and... | 9 | Given two integer arrays `startTime` and `endTime` and given an integer `queryTime`.
The `ith` student started doing their homework at the time `startTime[i]` and finished it at time `endTime[i]`.
Return _the number of students_ doing their homework at time `queryTime`. More formally, return the number of students wh... | Use the DFS to reconstruct the tree such that no leaf node is equal to the target. If the leaf node is equal to the target, return an empty object instead. |
Easy python solution 🎉 | number-of-students-doing-homework-at-a-given-time | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->First we run a loop till the length of any one the given array.Then as we know the length of both the arrays are equal so we check the elements of both array at once and... | 9 | Given an integer `n` and an integer array `rounds`. We have a circular track which consists of `n` sectors labeled from `1` to `n`. A marathon will be held on this track, the marathon consists of `m` rounds. The `ith` round starts at sector `rounds[i - 1]` and ends at sector `rounds[i]`. For example, round 1 starts at ... | Imagine that startTime[i] and endTime[i] form an interval (i.e. [startTime[i], endTime[i]]). The answer is how many times the queryTime laid in those mentioned intervals. |
Very Simple Python ||O(N) | number-of-students-doing-homework-at-a-given-time | 0 | 1 | Time Complexcity O(N)\nSpace Complexcity O(1)\n```\nclass Solution:\n def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n count=0\n n=len(endTime)\n for i in range(n):\n if endTime[i]>=queryTime and startTime[i]<=queryTime:\n count+... | 2 | Given two integer arrays `startTime` and `endTime` and given an integer `queryTime`.
The `ith` student started doing their homework at the time `startTime[i]` and finished it at time `endTime[i]`.
Return _the number of students_ doing their homework at time `queryTime`. More formally, return the number of students wh... | Use the DFS to reconstruct the tree such that no leaf node is equal to the target. If the leaf node is equal to the target, return an empty object instead. |
Very Simple Python ||O(N) | number-of-students-doing-homework-at-a-given-time | 0 | 1 | Time Complexcity O(N)\nSpace Complexcity O(1)\n```\nclass Solution:\n def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n count=0\n n=len(endTime)\n for i in range(n):\n if endTime[i]>=queryTime and startTime[i]<=queryTime:\n count+... | 2 | Given an integer `n` and an integer array `rounds`. We have a circular track which consists of `n` sectors labeled from `1` to `n`. A marathon will be held on this track, the marathon consists of `m` rounds. The `ith` round starts at sector `rounds[i - 1]` and ends at sector `rounds[i]`. For example, round 1 starts at ... | Imagine that startTime[i] and endTime[i] form an interval (i.e. [startTime[i], endTime[i]]). The answer is how many times the queryTime laid in those mentioned intervals. |
Python Simple Solution | Zip & Iterate - 86% 37ms | number-of-students-doing-homework-at-a-given-time | 0 | 1 | There are a lot of different ways to solve this problem, including a multitude of one liners, but I didn\'t see this method posted. Generally a fan of one liners as it helps condense code a bit, but there comes a point where they\'re *too* long.\n```\nclass Solution:\n def busyStudent(self, startTime: List[int], end... | 5 | Given two integer arrays `startTime` and `endTime` and given an integer `queryTime`.
The `ith` student started doing their homework at the time `startTime[i]` and finished it at time `endTime[i]`.
Return _the number of students_ doing their homework at time `queryTime`. More formally, return the number of students wh... | Use the DFS to reconstruct the tree such that no leaf node is equal to the target. If the leaf node is equal to the target, return an empty object instead. |
Python Simple Solution | Zip & Iterate - 86% 37ms | number-of-students-doing-homework-at-a-given-time | 0 | 1 | There are a lot of different ways to solve this problem, including a multitude of one liners, but I didn\'t see this method posted. Generally a fan of one liners as it helps condense code a bit, but there comes a point where they\'re *too* long.\n```\nclass Solution:\n def busyStudent(self, startTime: List[int], end... | 5 | Given an integer `n` and an integer array `rounds`. We have a circular track which consists of `n` sectors labeled from `1` to `n`. A marathon will be held on this track, the marathon consists of `m` rounds. The `ith` round starts at sector `rounds[i - 1]` and ends at sector `rounds[i]`. For example, round 1 starts at ... | Imagine that startTime[i] and endTime[i] form an interval (i.e. [startTime[i], endTime[i]]). The answer is how many times the queryTime laid in those mentioned intervals. |
Python One-Liner | number-of-students-doing-homework-at-a-given-time | 0 | 1 | ```\nclass Solution:\n def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n return sum([queryTime>=i and queryTime<=j for i, j in zip(startTime, endTime)])\n \n``` | 11 | Given two integer arrays `startTime` and `endTime` and given an integer `queryTime`.
The `ith` student started doing their homework at the time `startTime[i]` and finished it at time `endTime[i]`.
Return _the number of students_ doing their homework at time `queryTime`. More formally, return the number of students wh... | Use the DFS to reconstruct the tree such that no leaf node is equal to the target. If the leaf node is equal to the target, return an empty object instead. |
Python One-Liner | number-of-students-doing-homework-at-a-given-time | 0 | 1 | ```\nclass Solution:\n def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n return sum([queryTime>=i and queryTime<=j for i, j in zip(startTime, endTime)])\n \n``` | 11 | Given an integer `n` and an integer array `rounds`. We have a circular track which consists of `n` sectors labeled from `1` to `n`. A marathon will be held on this track, the marathon consists of `m` rounds. The `ith` round starts at sector `rounds[i - 1]` and ends at sector `rounds[i]`. For example, round 1 starts at ... | Imagine that startTime[i] and endTime[i] form an interval (i.e. [startTime[i], endTime[i]]). The answer is how many times the queryTime laid in those mentioned intervals. |
Python one line solution | number-of-students-doing-homework-at-a-given-time | 0 | 1 | **Python :**\n\n```\ndef busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n\treturn sum([i <= queryTime and queryTime <= j for i, j in zip(startTime, endTime)])\n```\n\n**Like it ? please upvote !** | 4 | Given two integer arrays `startTime` and `endTime` and given an integer `queryTime`.
The `ith` student started doing their homework at the time `startTime[i]` and finished it at time `endTime[i]`.
Return _the number of students_ doing their homework at time `queryTime`. More formally, return the number of students wh... | Use the DFS to reconstruct the tree such that no leaf node is equal to the target. If the leaf node is equal to the target, return an empty object instead. |
Python one line solution | number-of-students-doing-homework-at-a-given-time | 0 | 1 | **Python :**\n\n```\ndef busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n\treturn sum([i <= queryTime and queryTime <= j for i, j in zip(startTime, endTime)])\n```\n\n**Like it ? please upvote !** | 4 | Given an integer `n` and an integer array `rounds`. We have a circular track which consists of `n` sectors labeled from `1` to `n`. A marathon will be held on this track, the marathon consists of `m` rounds. The `ith` round starts at sector `rounds[i - 1]` and ends at sector `rounds[i]`. For example, round 1 starts at ... | Imagine that startTime[i] and endTime[i] form an interval (i.e. [startTime[i], endTime[i]]). The answer is how many times the queryTime laid in those mentioned intervals. |
Python simple solution | number-of-students-doing-homework-at-a-given-time | 0 | 1 | ```\nclass Solution:\n def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n ans = 0\n for i in range(len(startTime)):\n if startTime[i] <= queryTime <= endTime[i]:\n ans += 1\n return ans\n``` | 1 | Given two integer arrays `startTime` and `endTime` and given an integer `queryTime`.
The `ith` student started doing their homework at the time `startTime[i]` and finished it at time `endTime[i]`.
Return _the number of students_ doing their homework at time `queryTime`. More formally, return the number of students wh... | Use the DFS to reconstruct the tree such that no leaf node is equal to the target. If the leaf node is equal to the target, return an empty object instead. |
Python simple solution | number-of-students-doing-homework-at-a-given-time | 0 | 1 | ```\nclass Solution:\n def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n ans = 0\n for i in range(len(startTime)):\n if startTime[i] <= queryTime <= endTime[i]:\n ans += 1\n return ans\n``` | 1 | Given an integer `n` and an integer array `rounds`. We have a circular track which consists of `n` sectors labeled from `1` to `n`. A marathon will be held on this track, the marathon consists of `m` rounds. The `ith` round starts at sector `rounds[i - 1]` and ends at sector `rounds[i]`. For example, round 1 starts at ... | Imagine that startTime[i] and endTime[i] form an interval (i.e. [startTime[i], endTime[i]]). The answer is how many times the queryTime laid in those mentioned intervals. |
Simple python solution | number-of-students-doing-homework-at-a-given-time | 0 | 1 | ```\nclass Solution:\n def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n count = 0\n for i in range(len(startTime)):\n if queryTime <= endTime[i] and queryTime >= startTime[i]:\n count += 1\n return count\n``` | 2 | Given two integer arrays `startTime` and `endTime` and given an integer `queryTime`.
The `ith` student started doing their homework at the time `startTime[i]` and finished it at time `endTime[i]`.
Return _the number of students_ doing their homework at time `queryTime`. More formally, return the number of students wh... | Use the DFS to reconstruct the tree such that no leaf node is equal to the target. If the leaf node is equal to the target, return an empty object instead. |
Simple python solution | number-of-students-doing-homework-at-a-given-time | 0 | 1 | ```\nclass Solution:\n def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:\n count = 0\n for i in range(len(startTime)):\n if queryTime <= endTime[i] and queryTime >= startTime[i]:\n count += 1\n return count\n``` | 2 | Given an integer `n` and an integer array `rounds`. We have a circular track which consists of `n` sectors labeled from `1` to `n`. A marathon will be held on this track, the marathon consists of `m` rounds. The `ith` round starts at sector `rounds[i - 1]` and ends at sector `rounds[i]`. For example, round 1 starts at ... | Imagine that startTime[i] and endTime[i] form an interval (i.e. [startTime[i], endTime[i]]). The answer is how many times the queryTime laid in those mentioned intervals. |
[Python 3] Simple | Easy To Understand | Fast | rearrange-words-in-a-sentence | 0 | 1 | # Code\n```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n a = []\n for x in text.split(" "):\n a.append(x.lower())\n return " ".join(sorted(a, key=len)).capitalize()\n```\n\n# One Liner\n```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n retu... | 2 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
[Python 3] Simple | Easy To Understand | Fast | rearrange-words-in-a-sentence | 0 | 1 | # Code\n```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n a = []\n for x in text.split(" "):\n a.append(x.lower())\n return " ".join(sorted(a, key=len)).capitalize()\n```\n\n# One Liner\n```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n retu... | 2 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
[Python3] one line | rearrange-words-in-a-sentence | 0 | 1 | A few string operations chained together to get the job done. \n\n```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n return " ".join(sorted(text.split(), key=len)).capitalize()\n``` | 45 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
[Python3] one line | rearrange-words-in-a-sentence | 0 | 1 | A few string operations chained together to get the job done. \n\n```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n return " ".join(sorted(text.split(), key=len)).capitalize()\n``` | 45 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Python | Easy Solution✅ | rearrange-words-in-a-sentence | 0 | 1 | ```\ndef arrangeWords(self, text: str) -> str:\n seen = {}\n output = ""\n text_list = text.split()\n for word in text_list:\n n = len(word)\n if n in seen:\n value = seen[n]\n value.append(word.lower())\n seen[n] = value\n ... | 7 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Python | Easy Solution✅ | rearrange-words-in-a-sentence | 0 | 1 | ```\ndef arrangeWords(self, text: str) -> str:\n seen = {}\n output = ""\n text_list = text.split()\n for word in text_list:\n n = len(word)\n if n in seen:\n value = seen[n]\n value.append(word.lower())\n seen[n] = value\n ... | 7 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Python Easy Solution Using Dictionary and Sorting | rearrange-words-in-a-sentence | 0 | 1 | ```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n a=text.split()\n d,t={},\'\'\n for i in a:\n l=len(i)\n if l in d:\n d[l]+=" "+i\n else:\n d[l]=i\n for i in sorted(d):\n t+=" "+d[i]\n t=t.l... | 2 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Python Easy Solution Using Dictionary and Sorting | rearrange-words-in-a-sentence | 0 | 1 | ```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n a=text.split()\n d,t={},\'\'\n for i in a:\n l=len(i)\n if l in d:\n d[l]+=" "+i\n else:\n d[l]=i\n for i in sorted(d):\n t+=" "+d[i]\n t=t.l... | 2 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Python 3 -- Simple -- Readable -- Fast | rearrange-words-in-a-sentence | 0 | 1 | ```python\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n \n # convert string to an array\n arr = text.split()\n \n # convert the first letter of the array to a lower case letter\n arr[0] = arr[0].lower()\n \n # sort array by length\n arr.... | 8 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Python 3 -- Simple -- Readable -- Fast | rearrange-words-in-a-sentence | 0 | 1 | ```python\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n \n # convert string to an array\n arr = text.split()\n \n # convert the first letter of the array to a lower case letter\n arr[0] = arr[0].lower()\n \n # sort array by length\n arr.... | 8 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Simple and Understandable solution in Python | rearrange-words-in-a-sentence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Simple and Understandable solution in Python | rearrange-words-in-a-sentence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Python3 97 % | rearrange-words-in-a-sentence | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n text = text[0].lower()+text[1:]\n t = text.split()\n t.sort(key = len)\n t[0] = t[0].capitalize()\n return " ".join(t)\n\n\n \n``` | 0 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Python3 97 % | rearrange-words-in-a-sentence | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n text = text[0].lower()+text[1:]\n t = text.split()\n t.sort(key = len)\n t[0] = t[0].capitalize()\n return " ".join(t)\n\n\n \n``` | 0 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Python3 Solution (Simple | Fast | Efficient) -> ✅ | rearrange-words-in-a-sentence | 0 | 1 | # Approach (Step by step)\n1. Split text using **split()** method and store it inside the list:\n```python\nwords = text.split()\n```\n2. Make the first word lowercase using **lower()** method:\n```python\nwords[0] = words[0].lower()\n```\n3. Sort the list by length using **sorted()** function:\n```python\nwords = sort... | 0 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Python3 Solution (Simple | Fast | Efficient) -> ✅ | rearrange-words-in-a-sentence | 0 | 1 | # Approach (Step by step)\n1. Split text using **split()** method and store it inside the list:\n```python\nwords = text.split()\n```\n2. Make the first word lowercase using **lower()** method:\n```python\nwords[0] = words[0].lower()\n```\n3. Sort the list by length using **sorted()** function:\n```python\nwords = sort... | 0 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Beats 98.65% with SIMPLE Python solution. | rearrange-words-in-a-sentence | 0 | 1 | # Approach\nSplit the string into tokens. Only need to process the first token to be lower as the rest of the tokens will be the same. Sort by length. Then, make sure that the first string is capitalize correctly.\n\n# Complexity\n- Time complexity:\nO(n), where n is the length of the string\n\n- Space complexity:\nO(n... | 0 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Beats 98.65% with SIMPLE Python solution. | rearrange-words-in-a-sentence | 0 | 1 | # Approach\nSplit the string into tokens. Only need to process the first token to be lower as the rest of the tokens will be the same. Sort by length. Then, make sure that the first string is capitalize correctly.\n\n# Complexity\n- Time complexity:\nO(n), where n is the length of the string\n\n- Space complexity:\nO(n... | 0 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Python O(n) solution without sorting, readable and easy to understand. | rearrange-words-in-a-sentence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf we can group together all the words that have the same length, then we will be able to solve this problem. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Group the words that have the same length. \n2. Iter... | 0 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Python O(n) solution without sorting, readable and easy to understand. | rearrange-words-in-a-sentence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf we can group together all the words that have the same length, then we will be able to solve this problem. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Group the words that have the same length. \n2. Iter... | 0 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Simple python3 solution | Split + Sorting | rearrange-words-in-a-sentence | 0 | 1 | # Complexity\n- Time complexity: $$O(n \\cdot \\log(n))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n``` python3 []\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n data = [\n ... | 0 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Simple python3 solution | Split + Sorting | rearrange-words-in-a-sentence | 0 | 1 | # Complexity\n- Time complexity: $$O(n \\cdot \\log(n))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n``` python3 []\nclass Solution:\n def arrangeWords(self, text: str) -> str:\n data = [\n ... | 0 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Most Easiest and understandable python code | rearrange-words-in-a-sentence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given a sentence `text` (A _sentence_ is a string of space-separated words) in the following format:
* First letter is in upper case.
* Each word in `text` are separated by a single space.
Your task is to rearrange the words in text such that all words are rearranged in an increasing order of their lengths. If tw... | Create intervals of the area covered by each tap, sort intervals by the left end. We need to cover the interval [0, n]. we can start with the first interval and out of all intervals that intersect with it we choose the one that covers the farthest point to the right. What if there is a gap between intervals that is not... |
Most Easiest and understandable python code | rearrange-words-in-a-sentence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | There are `3n` piles of coins of varying size, you and your friends will take piles of coins as follows:
* In each step, you will choose **any** `3` piles of coins (not necessarily consecutive).
* Of your choice, Alice will pick the pile with the maximum number of coins.
* You will pick the next pile with the ma... | Store each word and their relative position. Then, sort them by length of words in case of tie by their original order. |
Python Simple Solution - Using Sets!! | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Complexity\n- Time complexity: O(N*N)\n- Space complexity: O(M)\n\n# Code\n```\nclass Solution:\n def peopleIndexes(self, favoriteCompanies: List[List[str]]) -> List[int]:\n \n answer: list = []\n favoriteCompanies = [set(companies) for companies in favoriteCompanies]\n\n for index, com... | 1 | Given the array `favoriteCompanies` where `favoriteCompanies[i]` is the list of favorites companies for the `ith` person (**indexed from 0**).
_Return the indices of people whose list of favorite companies is not a **subset** of any other list of favorites companies_. You must return the indices in increasing order.
... | null |
Python Simple Solution - Using Sets!! | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Complexity\n- Time complexity: O(N*N)\n- Space complexity: O(M)\n\n# Code\n```\nclass Solution:\n def peopleIndexes(self, favoriteCompanies: List[List[str]]) -> List[int]:\n \n answer: list = []\n favoriteCompanies = [set(companies) for companies in favoriteCompanies]\n\n for index, com... | 1 | Given an array `arr` that represents a permutation of numbers from `1` to `n`.
You have a binary string of size `n` that initially has all its bits set to zero. At each step `i` (assuming both the binary string and `arr` are 1-indexed) from `1` to `n`, the bit at position `arr[i]` is set to `1`.
You are also given an... | Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list. In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is ... |
Python | Dictionary + Bitwise operation | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAt first sight, the naive way to do it is to use two for loops + issubset. However, the name of company could be long, which makes issubset time consuming. \n\nSo how about replace the issubset by bitwise operation? \n# Approach\n<!-- Des... | 1 | Given the array `favoriteCompanies` where `favoriteCompanies[i]` is the list of favorites companies for the `ith` person (**indexed from 0**).
_Return the indices of people whose list of favorite companies is not a **subset** of any other list of favorites companies_. You must return the indices in increasing order.
... | null |
Python | Dictionary + Bitwise operation | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAt first sight, the naive way to do it is to use two for loops + issubset. However, the name of company could be long, which makes issubset time consuming. \n\nSo how about replace the issubset by bitwise operation? \n# Approach\n<!-- Des... | 1 | Given an array `arr` that represents a permutation of numbers from `1` to `n`.
You have a binary string of size `n` that initially has all its bits set to zero. At each step `i` (assuming both the binary string and `arr` are 1-indexed) from `1` to `n`, the bit at position `arr[i]` is set to `1`.
You are also given an... | Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list. In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is ... |
Python3 Clean Solution , Set , For Else | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def peopleIndexes(self, fc: List[List[str]]) -> List[int]:\n \n \n n=len(fc)\n ans=[]\n \n for i in range(n):\n s=set(fc[i])\n \n for j in range(n):\n if i==j:\n continue\n ... | 0 | Given the array `favoriteCompanies` where `favoriteCompanies[i]` is the list of favorites companies for the `ith` person (**indexed from 0**).
_Return the indices of people whose list of favorite companies is not a **subset** of any other list of favorites companies_. You must return the indices in increasing order.
... | null |
Python3 Clean Solution , Set , For Else | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def peopleIndexes(self, fc: List[List[str]]) -> List[int]:\n \n \n n=len(fc)\n ans=[]\n \n for i in range(n):\n s=set(fc[i])\n \n for j in range(n):\n if i==j:\n continue\n ... | 0 | Given an array `arr` that represents a permutation of numbers from `1` to `n`.
You have a binary string of size `n` that initially has all its bits set to zero. At each step `i` (assuming both the binary string and `arr` are 1-indexed) from `1` to `n`, the bit at position `arr[i]` is set to `1`.
You are also given an... | Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list. In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is ... |
Set Theory | Commented and Explained | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nA subset is a set that contains all components of a superset (and a superset has things a subset does not, as well as everything a subset to it has). \n\nDue to the nature of sub and super sets you\'ll likely run into the issue of memory ... | 0 | Given the array `favoriteCompanies` where `favoriteCompanies[i]` is the list of favorites companies for the `ith` person (**indexed from 0**).
_Return the indices of people whose list of favorite companies is not a **subset** of any other list of favorites companies_. You must return the indices in increasing order.
... | null |
Set Theory | Commented and Explained | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nA subset is a set that contains all components of a superset (and a superset has things a subset does not, as well as everything a subset to it has). \n\nDue to the nature of sub and super sets you\'ll likely run into the issue of memory ... | 0 | Given an array `arr` that represents a permutation of numbers from `1` to `n`.
You have a binary string of size `n` that initially has all its bits set to zero. At each step `i` (assuming both the binary string and `arr` are 1-indexed) from `1` to `n`, the bit at position `arr[i]` is set to `1`.
You are also given an... | Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list. In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is ... |
✅ The Fastest Approach Beats 100% | Explanation ✅ | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | \n\n\n# Approach\nThe main idea is to reduce nested loop time. In this case we iterates over all elements in the list that come **AFTER** the current element, instead of full iteration. It helps reduce time ... | 0 | Given the array `favoriteCompanies` where `favoriteCompanies[i]` is the list of favorites companies for the `ith` person (**indexed from 0**).
_Return the indices of people whose list of favorite companies is not a **subset** of any other list of favorites companies_. You must return the indices in increasing order.
... | null |
✅ The Fastest Approach Beats 100% | Explanation ✅ | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | \n\n\n# Approach\nThe main idea is to reduce nested loop time. In this case we iterates over all elements in the list that come **AFTER** the current element, instead of full iteration. It helps reduce time ... | 0 | Given an array `arr` that represents a permutation of numbers from `1` to `n`.
You have a binary string of size `n` that initially has all its bits set to zero. At each step `i` (assuming both the binary string and `arr` are 1-indexed) from `1` to `n`, the bit at position `arr[i]` is set to `1`.
You are also given an... | Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list. In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is ... |
Python solution, readable and beats 100% memory | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI just wanna solve stuff without the i,j and out of real world no library usages\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$... | 0 | Given the array `favoriteCompanies` where `favoriteCompanies[i]` is the list of favorites companies for the `ith` person (**indexed from 0**).
_Return the indices of people whose list of favorite companies is not a **subset** of any other list of favorites companies_. You must return the indices in increasing order.
... | null |
Python solution, readable and beats 100% memory | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI just wanna solve stuff without the i,j and out of real world no library usages\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$... | 0 | Given an array `arr` that represents a permutation of numbers from `1` to `n`.
You have a binary string of size `n` that initially has all its bits set to zero. At each step `i` (assuming both the binary string and `arr` are 1-indexed) from `1` to `n`, the bit at position `arr[i]` is set to `1`.
You are also given an... | Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list. In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is ... |
Python - 3 lines, simple with set difference | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\nConvert every list of strings to set of hash values of strings.\nDo index to index comaprison using set difference.\n\n# Complexity\n- Time complexity:\n$$O(n^2 F)$$ - F is length of list of favorites\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```\nclass Solution:\n def peopleIndexes(self, favoriteComp... | 0 | Given the array `favoriteCompanies` where `favoriteCompanies[i]` is the list of favorites companies for the `ith` person (**indexed from 0**).
_Return the indices of people whose list of favorite companies is not a **subset** of any other list of favorites companies_. You must return the indices in increasing order.
... | null |
Python - 3 lines, simple with set difference | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\nConvert every list of strings to set of hash values of strings.\nDo index to index comaprison using set difference.\n\n# Complexity\n- Time complexity:\n$$O(n^2 F)$$ - F is length of list of favorites\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```\nclass Solution:\n def peopleIndexes(self, favoriteComp... | 0 | Given an array `arr` that represents a permutation of numbers from `1` to `n`.
You have a binary string of size `n` that initially has all its bits set to zero. At each step `i` (assuming both the binary string and `arr` are 1-indexed) from `1` to `n`, the bit at position `arr[i]` is set to `1`.
You are also given an... | Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list. In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is ... |
bad ass fuck | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given the array `favoriteCompanies` where `favoriteCompanies[i]` is the list of favorites companies for the `ith` person (**indexed from 0**).
_Return the indices of people whose list of favorite companies is not a **subset** of any other list of favorites companies_. You must return the indices in increasing order.
... | null |
bad ass fuck | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given an array `arr` that represents a permutation of numbers from `1` to `n`.
You have a binary string of size `n` that initially has all its bits set to zero. At each step `i` (assuming both the binary string and `arr` are 1-indexed) from `1` to `n`, the bit at position `arr[i]` is set to `1`.
You are also given an... | Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list. In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is ... |
python super easy using set and map | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given the array `favoriteCompanies` where `favoriteCompanies[i]` is the list of favorites companies for the `ith` person (**indexed from 0**).
_Return the indices of people whose list of favorite companies is not a **subset** of any other list of favorites companies_. You must return the indices in increasing order.
... | null |
python super easy using set and map | people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Given an array `arr` that represents a permutation of numbers from `1` to `n`.
You have a binary string of size `n` that initially has all its bits set to zero. At each step `i` (assuming both the binary string and `arr` are 1-indexed) from `1` to `n`, the bit at position `arr[i]` is set to `1`.
You are also given an... | Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list. In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is ... |
[Python3] angular sweep O(N^2 logN) | maximum-number-of-darts-inside-of-a-circular-dartboard | 0 | 1 | Algo\nPick a point, say P, from the set, and rotate a circle with fixed-radius `r`. During the rotation P lies on the circumference of the circle (note P is not the center) and maintain a count of the number of points in the circle at an angle \u0398 (between PC and x-axis, where C is the center of the circle). \n\nFor... | 42 | Alice is throwing `n` darts on a very large wall. You are given an array `darts` where `darts[i] = [xi, yi]` is the position of the `ith` dart that Alice threw on the wall.
Bob knows the positions of the `n` darts on the wall. He wants to place a dartboard of radius `r` on the wall so that the maximum number of darts ... | null |
[Python3] angular sweep O(N^2 logN) | maximum-number-of-darts-inside-of-a-circular-dartboard | 0 | 1 | Algo\nPick a point, say P, from the set, and rotate a circle with fixed-radius `r`. During the rotation P lies on the circumference of the circle (note P is not the center) and maintain a count of the number of points in the circle at an angle \u0398 (between PC and x-axis, where C is the center of the circle). \n\nFor... | 42 | There are several stones **arranged in a row**, and each stone has an associated value which is an integer given in the array `stoneValue`.
In each round of the game, Alice divides the row into **two non-empty rows** (i.e. left row and right row), then Bob calculates the value of each row which is the sum of the value... | If there is an optimal solution, you can always move the circle so that two points lie on the boundary of the circle. When the radius is fixed, you can find either 0 or 1 or 2 circles that pass two given points at the same time. Loop for each pair of points and find the center of the circle, after that count the number... |
Track my solution | maximum-number-of-darts-inside-of-a-circular-dartboard | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | Alice is throwing `n` darts on a very large wall. You are given an array `darts` where `darts[i] = [xi, yi]` is the position of the `ith` dart that Alice threw on the wall.
Bob knows the positions of the `n` darts on the wall. He wants to place a dartboard of radius `r` on the wall so that the maximum number of darts ... | null |
Track my solution | maximum-number-of-darts-inside-of-a-circular-dartboard | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | There are several stones **arranged in a row**, and each stone has an associated value which is an integer given in the array `stoneValue`.
In each round of the game, Alice divides the row into **two non-empty rows** (i.e. left row and right row), then Bob calculates the value of each row which is the sum of the value... | If there is an optimal solution, you can always move the circle so that two points lie on the boundary of the circle. When the radius is fixed, you can find either 0 or 1 or 2 circles that pass two given points at the same time. Loop for each pair of points and find the center of the circle, after that count the number... |
Physical Simulation | Commented and Explained | O(P^2) Time and Space | maximum-number-of-darts-inside-of-a-circular-dartboard | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nRather than given the integer r for the radius of the circle, if we consider the points that Bob knows, we know that Bob can find the grouping (or clustering) of points that are within the best distance of the other points. This is a prob... | 0 | Alice is throwing `n` darts on a very large wall. You are given an array `darts` where `darts[i] = [xi, yi]` is the position of the `ith` dart that Alice threw on the wall.
Bob knows the positions of the `n` darts on the wall. He wants to place a dartboard of radius `r` on the wall so that the maximum number of darts ... | null |
Physical Simulation | Commented and Explained | O(P^2) Time and Space | maximum-number-of-darts-inside-of-a-circular-dartboard | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nRather than given the integer r for the radius of the circle, if we consider the points that Bob knows, we know that Bob can find the grouping (or clustering) of points that are within the best distance of the other points. This is a prob... | 0 | There are several stones **arranged in a row**, and each stone has an associated value which is an integer given in the array `stoneValue`.
In each round of the game, Alice divides the row into **two non-empty rows** (i.e. left row and right row), then Bob calculates the value of each row which is the sum of the value... | If there is an optimal solution, you can always move the circle so that two points lie on the boundary of the circle. When the radius is fixed, you can find either 0 or 1 or 2 circles that pass two given points at the same time. Loop for each pair of points and find the center of the circle, after that count the number... |
Clean Python solution with explaination | maximum-number-of-darts-inside-of-a-circular-dartboard | 0 | 1 | Step by step:\n1. The desired circle should be passed through at least 2 points out of those n points.\n\n2. Hence, let consider all pairs of those 2 points with the construction of the circle with radius r passes through those 2 points.\n\n3. Compute all possible centers (up to two). The trick in compute more faster i... | 4 | Alice is throwing `n` darts on a very large wall. You are given an array `darts` where `darts[i] = [xi, yi]` is the position of the `ith` dart that Alice threw on the wall.
Bob knows the positions of the `n` darts on the wall. He wants to place a dartboard of radius `r` on the wall so that the maximum number of darts ... | null |
Clean Python solution with explaination | maximum-number-of-darts-inside-of-a-circular-dartboard | 0 | 1 | Step by step:\n1. The desired circle should be passed through at least 2 points out of those n points.\n\n2. Hence, let consider all pairs of those 2 points with the construction of the circle with radius r passes through those 2 points.\n\n3. Compute all possible centers (up to two). The trick in compute more faster i... | 4 | There are several stones **arranged in a row**, and each stone has an associated value which is an integer given in the array `stoneValue`.
In each round of the game, Alice divides the row into **two non-empty rows** (i.e. left row and right row), then Bob calculates the value of each row which is the sum of the value... | If there is an optimal solution, you can always move the circle so that two points lie on the boundary of the circle. When the radius is fixed, you can find either 0 or 1 or 2 circles that pass two given points at the same time. Loop for each pair of points and find the center of the circle, after that count the number... |
Simple Python solution using startswith | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def isPrefixOfWord(self, s: str, searchWord: str) -> int:\n a = s.split()\n for i in range(len(a)):\n if a[i].startswith(searchWord):\n return i+1\n return -1\n``` | 2 | Given a `sentence` that consists of some words separated by a **single space**, and a `searchWord`, check if `searchWord` is a prefix of any word in `sentence`.
Return _the index of the word in_ `sentence` _(**1-indexed**) where_ `searchWord` _is a prefix of this word_. If `searchWord` is a prefix of more than one wor... | Do the filtering and sort as said. Note that the id may not be the index in the array. |
Simple Python solution using startswith | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def isPrefixOfWord(self, s: str, searchWord: str) -> int:\n a = s.split()\n for i in range(len(a)):\n if a[i].startswith(searchWord):\n return i+1\n return -1\n``` | 2 | Given an array of positive integers `arr`, find a pattern of length `m` that is repeated `k` or more times.
A **pattern** is a subarray (consecutive sub-sequence) that consists of one or more values, repeated multiple times **consecutively** without overlapping. A pattern is defined by its length and the number of rep... | First extract the words of the sentence. Check for each word if searchWord occurs at index 0, if so return the index of this word (1-indexed) If searchWord doesn't exist as a prefix of any word return the default value (-1). |
python3 code | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | # Code\n```\nclass Solution:\n def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:\n words = sentence.split()\n indicies = []\n for i, word in enumerate(words):\n if word.startswith(searchWord):\n return i + 1\n return -1\n```\n\n where_ `searchWord` _is a prefix of this word_. If `searchWord` is a prefix of more than one wor... | Do the filtering and sort as said. Note that the id may not be the index in the array. |
python3 code | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | # Code\n```\nclass Solution:\n def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:\n words = sentence.split()\n indicies = []\n for i, word in enumerate(words):\n if word.startswith(searchWord):\n return i + 1\n return -1\n```\n\n that consists of one or more values, repeated multiple times **consecutively** without overlapping. A pattern is defined by its length and the number of rep... | First extract the words of the sentence. Check for each word if searchWord occurs at index 0, if so return the index of this word (1-indexed) If searchWord doesn't exist as a prefix of any word return the default value (-1). |
python3 code | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | # Code\n```\nclass Solution:\n def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:\n words = sentence.split()\n indicies = []\n for word in words:\n if word.startswith(searchWord):\n indicies.append(words.index(word))\n if len(indicies) > 0:\n ... | 1 | Given a `sentence` that consists of some words separated by a **single space**, and a `searchWord`, check if `searchWord` is a prefix of any word in `sentence`.
Return _the index of the word in_ `sentence` _(**1-indexed**) where_ `searchWord` _is a prefix of this word_. If `searchWord` is a prefix of more than one wor... | Do the filtering and sort as said. Note that the id may not be the index in the array. |
python3 code | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | # Code\n```\nclass Solution:\n def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:\n words = sentence.split()\n indicies = []\n for word in words:\n if word.startswith(searchWord):\n indicies.append(words.index(word))\n if len(indicies) > 0:\n ... | 1 | Given an array of positive integers `arr`, find a pattern of length `m` that is repeated `k` or more times.
A **pattern** is a subarray (consecutive sub-sequence) that consists of one or more values, repeated multiple times **consecutively** without overlapping. A pattern is defined by its length and the number of rep... | First extract the words of the sentence. Check for each word if searchWord occurs at index 0, if so return the index of this word (1-indexed) If searchWord doesn't exist as a prefix of any word return the default value (-1). |
Python with str.split(), 12ms | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | \n\t\n\tdef isPrefixOfWord(self, sentence, searchWord):\n sentence = sentence.split(\' \')\n for index,word in enumerate(sentence):\n if searchWord == word[:len(searchWord)]:\n return index+1\n return -1 | 13 | Given a `sentence` that consists of some words separated by a **single space**, and a `searchWord`, check if `searchWord` is a prefix of any word in `sentence`.
Return _the index of the word in_ `sentence` _(**1-indexed**) where_ `searchWord` _is a prefix of this word_. If `searchWord` is a prefix of more than one wor... | Do the filtering and sort as said. Note that the id may not be the index in the array. |
Python with str.split(), 12ms | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | \n\t\n\tdef isPrefixOfWord(self, sentence, searchWord):\n sentence = sentence.split(\' \')\n for index,word in enumerate(sentence):\n if searchWord == word[:len(searchWord)]:\n return index+1\n return -1 | 13 | Given an array of positive integers `arr`, find a pattern of length `m` that is repeated `k` or more times.
A **pattern** is a subarray (consecutive sub-sequence) that consists of one or more values, repeated multiple times **consecutively** without overlapping. A pattern is defined by its length and the number of rep... | First extract the words of the sentence. Check for each word if searchWord occurs at index 0, if so return the index of this word (1-indexed) If searchWord doesn't exist as a prefix of any word return the default value (-1). |
Fastest 99.85% solution & Beginner friendly... Easy and simple python solution | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | # Approach\nJust search for the given searchWord in each of the word of the sentence by checking beginning of the word till the length of the searchWord.. As for loop starts from 0, return the index incremented by 1 in order to get the position of the word in the sentence.\nSimple............!\n<!-- Describe your appro... | 1 | Given a `sentence` that consists of some words separated by a **single space**, and a `searchWord`, check if `searchWord` is a prefix of any word in `sentence`.
Return _the index of the word in_ `sentence` _(**1-indexed**) where_ `searchWord` _is a prefix of this word_. If `searchWord` is a prefix of more than one wor... | Do the filtering and sort as said. Note that the id may not be the index in the array. |
Fastest 99.85% solution & Beginner friendly... Easy and simple python solution | check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence | 0 | 1 | # Approach\nJust search for the given searchWord in each of the word of the sentence by checking beginning of the word till the length of the searchWord.. As for loop starts from 0, return the index incremented by 1 in order to get the position of the word in the sentence.\nSimple............!\n<!-- Describe your appro... | 1 | Given an array of positive integers `arr`, find a pattern of length `m` that is repeated `k` or more times.
A **pattern** is a subarray (consecutive sub-sequence) that consists of one or more values, repeated multiple times **consecutively** without overlapping. A pattern is defined by its length and the number of rep... | First extract the words of the sentence. Check for each word if searchWord occurs at index 0, if so return the index of this word (1-indexed) If searchWord doesn't exist as a prefix of any word return the default value (-1). |
🥇 EASY || JAVA || C++ || PYTHON || EXPLAINED WITH VIDEO || BEGINNER FRIENDLY 🔥🔥 | pseudo-palindromic-paths-in-a-binary-tree | 1 | 1 | **PLEASE LIKE AND COMMENT IF THIS SOLUTION HAS HELPED YOU**\n\n**VIDEO EXPLANATION OF SOLUTION:**\nhttps://www.youtube.com/watch?v=spC68MStRRg&ab_channel=AlgoLock\n\n**SOLUTION: JAVA**\n```java\nclass Solution {\n public int pseudoPalindromicPaths (TreeNode root) {\n return parsePalinTree(root, new int[10]);\... | 8 | Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be **pseudo-palindromic** if at least one permutation of the node values in the path is a palindrome.
_Return the number of **pseudo-palindromic** paths going from the root node to leaf nodes._
**Example 1:**
**Input:*... | Use DP. Try to cut the array into d non-empty sub-arrays. Try all possible cuts for the array. Use dp[i][j] where DP states are i the index of the last cut and j the number of remaining cuts. Complexity is O(n * n * d). |
🥇 EASY || JAVA || C++ || PYTHON || EXPLAINED WITH VIDEO || BEGINNER FRIENDLY 🔥🔥 | pseudo-palindromic-paths-in-a-binary-tree | 1 | 1 | **PLEASE LIKE AND COMMENT IF THIS SOLUTION HAS HELPED YOU**\n\n**VIDEO EXPLANATION OF SOLUTION:**\nhttps://www.youtube.com/watch?v=spC68MStRRg&ab_channel=AlgoLock\n\n**SOLUTION: JAVA**\n```java\nclass Solution {\n public int pseudoPalindromicPaths (TreeNode root) {\n return parsePalinTree(root, new int[10]);\... | 8 | You are given an `m x n` binary grid `grid` where `1` represents land and `0` represents water. An **island** is a maximal **4-directionally** (horizontal or vertical) connected group of `1`'s.
The grid is said to be **connected** if we have **exactly one island**, otherwise is said **disconnected**.
In one day, we a... | Note that the node values of a path form a palindrome if at most one digit has an odd frequency (parity). Use a Depth First Search (DFS) keeping the frequency (parity) of the digits. Once you are in a leaf node check if at most one digit has an odd frequency (parity). |
🔥 [LeetCode The Hard Way] 🔥 Explained Line By Line | pseudo-palindromic-paths-in-a-binary-tree | 0 | 1 | Please check out [LeetCode The Hard Way](https://wingkwong.github.io/leetcode-the-hard-way/) for more solution explanations and tutorials. \nI\'ll explain my solution line by line daily and you can find the full list in my [Discord](https://discord.gg/Nqm4jJcyBf).\nIf you like it, please give a star, watch my [Github R... | 78 | Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be **pseudo-palindromic** if at least one permutation of the node values in the path is a palindrome.
_Return the number of **pseudo-palindromic** paths going from the root node to leaf nodes._
**Example 1:**
**Input:*... | Use DP. Try to cut the array into d non-empty sub-arrays. Try all possible cuts for the array. Use dp[i][j] where DP states are i the index of the last cut and j the number of remaining cuts. Complexity is O(n * n * d). |
🔥 [LeetCode The Hard Way] 🔥 Explained Line By Line | pseudo-palindromic-paths-in-a-binary-tree | 0 | 1 | Please check out [LeetCode The Hard Way](https://wingkwong.github.io/leetcode-the-hard-way/) for more solution explanations and tutorials. \nI\'ll explain my solution line by line daily and you can find the full list in my [Discord](https://discord.gg/Nqm4jJcyBf).\nIf you like it, please give a star, watch my [Github R... | 78 | You are given an `m x n` binary grid `grid` where `1` represents land and `0` represents water. An **island** is a maximal **4-directionally** (horizontal or vertical) connected group of `1`'s.
The grid is said to be **connected** if we have **exactly one island**, otherwise is said **disconnected**.
In one day, we a... | Note that the node values of a path form a palindrome if at most one digit has an odd frequency (parity). Use a Depth First Search (DFS) keeping the frequency (parity) of the digits. Once you are in a leaf node check if at most one digit has an odd frequency (parity). |
Python | DFS & Set | With Explanation | Easy to Understand | pseudo-palindromic-paths-in-a-binary-tree | 0 | 1 | ```\n# Definition for a binary tree node.\n# class TreeNode:\n# def __init__(self, val=0, left=None, right=None):\n# self.val = val\n# self.left = left\n# self.right = right\nclass Solution:\n def pseudoPalindromicPaths (self, root: Optional[TreeNode]) -> int:\n # traverse the tree... | 56 | Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be **pseudo-palindromic** if at least one permutation of the node values in the path is a palindrome.
_Return the number of **pseudo-palindromic** paths going from the root node to leaf nodes._
**Example 1:**
**Input:*... | Use DP. Try to cut the array into d non-empty sub-arrays. Try all possible cuts for the array. Use dp[i][j] where DP states are i the index of the last cut and j the number of remaining cuts. Complexity is O(n * n * d). |
Python | DFS & Set | With Explanation | Easy to Understand | pseudo-palindromic-paths-in-a-binary-tree | 0 | 1 | ```\n# Definition for a binary tree node.\n# class TreeNode:\n# def __init__(self, val=0, left=None, right=None):\n# self.val = val\n# self.left = left\n# self.right = right\nclass Solution:\n def pseudoPalindromicPaths (self, root: Optional[TreeNode]) -> int:\n # traverse the tree... | 56 | You are given an `m x n` binary grid `grid` where `1` represents land and `0` represents water. An **island** is a maximal **4-directionally** (horizontal or vertical) connected group of `1`'s.
The grid is said to be **connected** if we have **exactly one island**, otherwise is said **disconnected**.
In one day, we a... | Note that the node values of a path form a palindrome if at most one digit has an odd frequency (parity). Use a Depth First Search (DFS) keeping the frequency (parity) of the digits. Once you are in a leaf node check if at most one digit has an odd frequency (parity). |
Preorder Traversal python solution using mapping | pseudo-palindromic-paths-in-a-binary-tree | 0 | 1 | ```\nclass Solution:\n def pseudoPalindromicPaths (self, root: Optional[TreeNode]) -> int:\n def pre(root,dic):\n if root is None:\n return 0\n if root.left is None and root.right is None:\n if root.val not in dic:\n dic[root.val]=1\n ... | 3 | Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be **pseudo-palindromic** if at least one permutation of the node values in the path is a palindrome.
_Return the number of **pseudo-palindromic** paths going from the root node to leaf nodes._
**Example 1:**
**Input:*... | Use DP. Try to cut the array into d non-empty sub-arrays. Try all possible cuts for the array. Use dp[i][j] where DP states are i the index of the last cut and j the number of remaining cuts. Complexity is O(n * n * d). |
Preorder Traversal python solution using mapping | pseudo-palindromic-paths-in-a-binary-tree | 0 | 1 | ```\nclass Solution:\n def pseudoPalindromicPaths (self, root: Optional[TreeNode]) -> int:\n def pre(root,dic):\n if root is None:\n return 0\n if root.left is None and root.right is None:\n if root.val not in dic:\n dic[root.val]=1\n ... | 3 | You are given an `m x n` binary grid `grid` where `1` represents land and `0` represents water. An **island** is a maximal **4-directionally** (horizontal or vertical) connected group of `1`'s.
The grid is said to be **connected** if we have **exactly one island**, otherwise is said **disconnected**.
In one day, we a... | Note that the node values of a path form a palindrome if at most one digit has an odd frequency (parity). Use a Depth First Search (DFS) keeping the frequency (parity) of the digits. Once you are in a leaf node check if at most one digit has an odd frequency (parity). |
✅ 90.14% Dynamic Programming Optimized | max-dot-product-of-two-subsequences | 1 | 1 | # Intuition\nWhen faced with two sequences and asked to find the maximum dot product, it becomes evident that we might have to consider various combinations of subsequences to determine the maximum value. This naturally hints towards the possibility of using dynamic programming, as there are overlapping subproblems.\n\... | 45 | Given two arrays `nums1` and `nums2`.
Return the maximum dot product between **non-empty** subsequences of nums1 and nums2 with the same length.
A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of... | Simulate the problem. Count the number of 1's in the binary representation of each integer. Sort by the number of 1's ascending and by the value in case of tie. |
✅ 90.14% Dynamic Programming Optimized | max-dot-product-of-two-subsequences | 1 | 1 | # Intuition\nWhen faced with two sequences and asked to find the maximum dot product, it becomes evident that we might have to consider various combinations of subsequences to determine the maximum value. This naturally hints towards the possibility of using dynamic programming, as there are overlapping subproblems.\n\... | 45 | Given an array `nums` that represents a permutation of integers from `1` to `n`. We are going to construct a binary search tree (BST) by inserting the elements of `nums` in order into an initially empty BST. Find the number of different ways to reorder `nums` so that the constructed BST is identical to that formed from... | Use dynamic programming, define DP[i][j] as the maximum dot product of two subsequences starting in the position i of nums1 and position j of nums2. |
【Video】Give me 10 minutes - How we think about a solution - Python, JavaScript, Java, C++ | max-dot-product-of-two-subsequences | 1 | 1 | Welcome to my post! This post starts with "How we think about a solution". In other words, that is my thought process to solve the question. This post explains how I get to my solution instead of just posting solution codes or out of blue algorithms. I hope it is helpful for someone.\n\n# Intuition\nUsing Dynamic Progr... | 37 | Given two arrays `nums1` and `nums2`.
Return the maximum dot product between **non-empty** subsequences of nums1 and nums2 with the same length.
A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of... | Simulate the problem. Count the number of 1's in the binary representation of each integer. Sort by the number of 1's ascending and by the value in case of tie. |
【Video】Give me 10 minutes - How we think about a solution - Python, JavaScript, Java, C++ | max-dot-product-of-two-subsequences | 1 | 1 | Welcome to my post! This post starts with "How we think about a solution". In other words, that is my thought process to solve the question. This post explains how I get to my solution instead of just posting solution codes or out of blue algorithms. I hope it is helpful for someone.\n\n# Intuition\nUsing Dynamic Progr... | 37 | Given an array `nums` that represents a permutation of integers from `1` to `n`. We are going to construct a binary search tree (BST) by inserting the elements of `nums` in order into an initially empty BST. Find the number of different ways to reorder `nums` so that the constructed BST is identical to that formed from... | Use dynamic programming, define DP[i][j] as the maximum dot product of two subsequences starting in the position i of nums1 and position j of nums2. |
🚀92.14% || DP Recursive & Iterative || Explained Intuition🚀 | max-dot-product-of-two-subsequences | 1 | 1 | # Porblem Description\n\nGiven two arrays, `nums1` and `nums2`. The task is to find the **maximum** dot product that can be obtained by taking **non-empty** **subsequences** of equal length from `nums1` and `nums2`.\n\nA **subsequence** of an array is a new array obtained by deleting some (or none) of the elements from... | 42 | Given two arrays `nums1` and `nums2`.
Return the maximum dot product between **non-empty** subsequences of nums1 and nums2 with the same length.
A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of... | Simulate the problem. Count the number of 1's in the binary representation of each integer. Sort by the number of 1's ascending and by the value in case of tie. |
🚀92.14% || DP Recursive & Iterative || Explained Intuition🚀 | max-dot-product-of-two-subsequences | 1 | 1 | # Porblem Description\n\nGiven two arrays, `nums1` and `nums2`. The task is to find the **maximum** dot product that can be obtained by taking **non-empty** **subsequences** of equal length from `nums1` and `nums2`.\n\nA **subsequence** of an array is a new array obtained by deleting some (or none) of the elements from... | 42 | Given an array `nums` that represents a permutation of integers from `1` to `n`. We are going to construct a binary search tree (BST) by inserting the elements of `nums` in order into an initially empty BST. Find the number of different ways to reorder `nums` so that the constructed BST is identical to that formed from... | Use dynamic programming, define DP[i][j] as the maximum dot product of two subsequences starting in the position i of nums1 and position j of nums2. |
✅☑[C++/C/Java/Python/JavaScript] || DP || Recursion || EXPLAINED🔥 | max-dot-product-of-two-subsequences | 1 | 1 | # *PLEASE UPVOTE IF IT HELPED*\n\n---\n\n\n# Approach\n***(Also explained in the code)***\n\n**Explanation:**\n\nThe problem can be solved using dynamic programming. The goal is to find the maximum dot product between two subsequences of the given arrays `nums1` and `nums2`. We can create a 2D DP array `dp`, where `dp[... | 2 | Given two arrays `nums1` and `nums2`.
Return the maximum dot product between **non-empty** subsequences of nums1 and nums2 with the same length.
A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of... | Simulate the problem. Count the number of 1's in the binary representation of each integer. Sort by the number of 1's ascending and by the value in case of tie. |
✅☑[C++/C/Java/Python/JavaScript] || DP || Recursion || EXPLAINED🔥 | max-dot-product-of-two-subsequences | 1 | 1 | # *PLEASE UPVOTE IF IT HELPED*\n\n---\n\n\n# Approach\n***(Also explained in the code)***\n\n**Explanation:**\n\nThe problem can be solved using dynamic programming. The goal is to find the maximum dot product between two subsequences of the given arrays `nums1` and `nums2`. We can create a 2D DP array `dp`, where `dp[... | 2 | Given an array `nums` that represents a permutation of integers from `1` to `n`. We are going to construct a binary search tree (BST) by inserting the elements of `nums` in order into an initially empty BST. Find the number of different ways to reorder `nums` so that the constructed BST is identical to that formed from... | Use dynamic programming, define DP[i][j] as the maximum dot product of two subsequences starting in the position i of nums1 and position j of nums2. |
EASY PYTHON SOLUTION USING DP | max-dot-product-of-two-subsequences | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def dp(self,i,j,nums1,nums2,ne,dct):\n if i==0 and j==0:\n if ne:\n return max(0,nums1[0]*nums2[0])\n return nums1[0]*nums2[0]\n if i<0 or j<0:\n if ne:\n return 0\n return float("-infinity")\n ... | 2 | Given two arrays `nums1` and `nums2`.
Return the maximum dot product between **non-empty** subsequences of nums1 and nums2 with the same length.
A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of... | Simulate the problem. Count the number of 1's in the binary representation of each integer. Sort by the number of 1's ascending and by the value in case of tie. |
EASY PYTHON SOLUTION USING DP | max-dot-product-of-two-subsequences | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def dp(self,i,j,nums1,nums2,ne,dct):\n if i==0 and j==0:\n if ne:\n return max(0,nums1[0]*nums2[0])\n return nums1[0]*nums2[0]\n if i<0 or j<0:\n if ne:\n return 0\n return float("-infinity")\n ... | 2 | Given an array `nums` that represents a permutation of integers from `1` to `n`. We are going to construct a binary search tree (BST) by inserting the elements of `nums` in order into an initially empty BST. Find the number of different ways to reorder `nums` so that the constructed BST is identical to that formed from... | Use dynamic programming, define DP[i][j] as the maximum dot product of two subsequences starting in the position i of nums1 and position j of nums2. |
Bottom Up Python Solution | max-dot-product-of-two-subsequences | 0 | 1 | # Code\n```\nclass Solution:\n def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:\n \'\'\'\n Recurrence can be seen as dp[i][j] = max(x, x + dp[i-1][j-1], dp[i][j-1], dp[i-1][j])\n \n Where:\n 1. x: This is the product of the current elements nums1[j-1] and nums2[i... | 1 | Given two arrays `nums1` and `nums2`.
Return the maximum dot product between **non-empty** subsequences of nums1 and nums2 with the same length.
A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of... | Simulate the problem. Count the number of 1's in the binary representation of each integer. Sort by the number of 1's ascending and by the value in case of tie. |
Bottom Up Python Solution | max-dot-product-of-two-subsequences | 0 | 1 | # Code\n```\nclass Solution:\n def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:\n \'\'\'\n Recurrence can be seen as dp[i][j] = max(x, x + dp[i-1][j-1], dp[i][j-1], dp[i-1][j])\n \n Where:\n 1. x: This is the product of the current elements nums1[j-1] and nums2[i... | 1 | Given an array `nums` that represents a permutation of integers from `1` to `n`. We are going to construct a binary search tree (BST) by inserting the elements of `nums` in order into an initially empty BST. Find the number of different ways to reorder `nums` so that the constructed BST is identical to that formed from... | Use dynamic programming, define DP[i][j] as the maximum dot product of two subsequences starting in the position i of nums1 and position j of nums2. |
[Python] Simplest Memoization DP Code with Approach | max-dot-product-of-two-subsequences | 0 | 1 | ### Approach:\n\n1. Base Condition - When either of the array elements are exhausted \n2. The Memoization `currentKey` should be made up of currentIndexes of both arrays\n3. Take into account all the three recursive calls:\n\ti. When both the indexes are incremented\n\tii. Two cases where one of the both indexes are in... | 1 | Given two arrays `nums1` and `nums2`.
Return the maximum dot product between **non-empty** subsequences of nums1 and nums2 with the same length.
A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of... | Simulate the problem. Count the number of 1's in the binary representation of each integer. Sort by the number of 1's ascending and by the value in case of tie. |
[Python] Simplest Memoization DP Code with Approach | max-dot-product-of-two-subsequences | 0 | 1 | ### Approach:\n\n1. Base Condition - When either of the array elements are exhausted \n2. The Memoization `currentKey` should be made up of currentIndexes of both arrays\n3. Take into account all the three recursive calls:\n\ti. When both the indexes are incremented\n\tii. Two cases where one of the both indexes are in... | 1 | Given an array `nums` that represents a permutation of integers from `1` to `n`. We are going to construct a binary search tree (BST) by inserting the elements of `nums` in order into an initially empty BST. Find the number of different ways to reorder `nums` so that the constructed BST is identical to that formed from... | Use dynamic programming, define DP[i][j] as the maximum dot product of two subsequences starting in the position i of nums1 and position j of nums2. |
Simple solution with HashMap in Python3 | make-two-arrays-equal-by-reversing-subarrays | 0 | 1 | # Intuition\nHere we have:\n- `target` and `arr` lists of integers\n- our goal is to check if we could reverse `arr` to make it as `target`\n\nAn approach is **extremely** simple:\n- create a **HashMap**, that\'ll represent a **balance** of frequencies of integers (**FOI**)\n- if FOI of each lists are **equal, then we ... | 1 | You are given two integer arrays of equal length `target` and `arr`. In one step, you can select any **non-empty subarray** of `arr` and reverse it. You are allowed to make any number of steps.
Return `true` _if you can make_ `arr` _equal to_ `target` _or_ `false` _otherwise_.
**Example 1:**
**Input:** target = \[1,... | For each position we simply need to find the first occurrence of a/b/c on or after this position. So we can pre-compute three link-list of indices of each a, b, and c. |
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