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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✅👇3 Approach🔥||✅👇Transitioning from a 🔥brute-force to optimized ones✅👇 | find-the-duplicate-number | 1 | 1 | # Problem Understanding\nGiven an array of N + 1 integers where each element is between 1 and N, the task is to find the duplicate number efficiently.\n\n# Approach\n\n---\n\n**I have explored three different approaches to solve this problem, and I am gradually transitioning from a brute-force approach to more optimize... | 60 | Given an array of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.
There is only **one repeated number** in `nums`, return _this repeated number_.
You must solve the problem **without** modifying the array `nums` and uses only constant extra space.
**Example 1:**
**... | null |
Easy python solution with O(n) time complexity with EXPLANANTION | find-the-duplicate-number | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt is given that the array consists integers from 1 to n only. So our task is easy we can just have an array of size n intialized with zeros.\n\nNow we iterate with say i throughtout the list and increment the i^th index of the array.\n\n... | 0 | Given an array of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.
There is only **one repeated number** in `nums`, return _this repeated number_.
You must solve the problem **without** modifying the array `nums` and uses only constant extra space.
**Example 1:**
**... | null |
Find the duplicate number | find-the-duplicate-number | 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 of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.
There is only **one repeated number** in `nums`, return _this repeated number_.
You must solve the problem **without** modifying the array `nums` and uses only constant extra space.
**Example 1:**
**... | null |
✅98.89%🔥Easy Solution🔥1 line code🔥 | find-the-duplicate-number | 0 | 1 | \n# Solution\n```\nclass Solution:\n def findDuplicate(self, nums: List[int]) -> int:\n seen = set()\n return next(num for num in nums if num in seen or seen.add(num))\n\n\n``` | 19 | Given an array of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.
There is only **one repeated number** in `nums`, return _this repeated number_.
You must solve the problem **without** modifying the array `nums` and uses only constant extra space.
**Example 1:**
**... | null |
Easy & Fastest - slow and fast pointer approach | find-the-duplicate-number | 0 | 1 | \n# Code\n```\nclass Solution:\n def findDuplicate(self, nums: List[int]) -> int:\n slow = nums[0]\n fast = nums[0]\n while True:\n slow = nums[slow]\n fast= nums[nums[fast]]\n if slow == fast: break\n \n fast = nums[0]\n while slow != fas... | 5 | Given an array of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.
There is only **one repeated number** in `nums`, return _this repeated number_.
You must solve the problem **without** modifying the array `nums` and uses only constant extra space.
**Example 1:**
**... | null |
Python | Using two pointers | find-the-duplicate-number | 0 | 1 | # Intuition\nUsing Floyd\'s and Hare algorithm for cycle detection in a linked list using 2 pointers `slow` and `fast` pointers.\n\n# Approach\n\nThis algorithm works by first finding the intersection point of the two pointers within the cycle (Phase 1), and then finding the entrance to the cycle (Phase 2). The entranc... | 1 | Given an array of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.
There is only **one repeated number** in `nums`, return _this repeated number_.
You must solve the problem **without** modifying the array `nums` and uses only constant extra space.
**Example 1:**
**... | null |
Using auxiliary board | Python | O(mn) | game-of-life | 0 | 1 | # Code\n```\nclass Solution:\n def gameOfLife(self, board: List[List[int]]) -> None:\n """\n Do not return anything, modify board in-place instead.\n """\n def withinbounds(a, b):\n if (a>=0 and a<=len(board)-1) and (b>=0 and b<=len(board[0])-1):\n return True\n ... | 2 | According to [Wikipedia's article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life): "The **Game of Life**, also known simply as **Life**, is a cellular automaton devised by the British mathematician John Horton Conway in 1970. "
The board is made up of an `m x n` grid of cells, where each cell has an initial st... | null |
289: Solution with step by step explanation | game-of-life | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe code implements the Game of Life, a cellular automaton, where each cell has an initial state: live (represented by a 1) or dead (represented by a 0). The function takes the current state of the board as input and returns... | 5 | According to [Wikipedia's article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life): "The **Game of Life**, also known simply as **Life**, is a cellular automaton devised by the British mathematician John Horton Conway in 1970. "
The board is made up of an `m x n` grid of cells, where each cell has an initial st... | null |
[Python] Very Simple Solution with Explanation, No space used | game-of-life | 0 | 1 | ```\nclass Solution:\n def gameOfLife(self, board: List[List[int]]) -> None:\n """\n Do not return anything, modify board in-place instead.\n """\n ## RC ##\n ## APPRAOCH : IN-PLACE MANIPULATION ##\n # when the value needs to be updated, we donot just change 0 to 1 / 1 to ... | 30 | According to [Wikipedia's article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life): "The **Game of Life**, also known simply as **Life**, is a cellular automaton devised by the British mathematician John Horton Conway in 1970. "
The board is made up of an `m x n` grid of cells, where each cell has an initial st... | null |
C++ and Python Simple Solutions w/comments, O(nm) time, O(1) Space | game-of-life | 0 | 1 | **C++:**\n```\nclass Solution {\npublic:\n // Helper function to check validility of neighbor\n bool isValidNeighbor(int x, int y, vector<vector<int>>& board) {\n return (x >= 0 && x < board.size() && y >= 0 && y < board[0].size());\n }\n \n void gameOfLife(vector<vector<int>>& board) {\n /... | 18 | According to [Wikipedia's article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life): "The **Game of Life**, also known simply as **Life**, is a cellular automaton devised by the British mathematician John Horton Conway in 1970. "
The board is made up of an `m x n` grid of cells, where each cell has an initial st... | null |
game of Life || Python3 || Arrays | game-of-life | 0 | 1 | ```\nclass Solution:\n def gameOfLife(self, board: List[List[int]]) -> None:\n """\n Do not return anything, modify board in-place instead.\n """\n dirs = [[1, 0], [0, 1], [-1, 0], [0, -1], [1, 1], [-1, -1], [1, -1], [-1, 1]]\n \n for i in range(0, len(board)):\n ... | 1 | According to [Wikipedia's article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life): "The **Game of Life**, also known simply as **Life**, is a cellular automaton devised by the British mathematician John Horton Conway in 1970. "
The board is made up of an `m x n` grid of cells, where each cell has an initial st... | null |
Python 3 || 2 lines, w/ explanation || T/M: 98% / 100% | word-pattern | 0 | 1 | A *bijection* is both *onto* and *one-to-one*. The figure below illustrates:\n- \n\n\nThese conditions for bijectivity are satisfied if and only if the following is true:\n- The counts of distinct eleme... | 161 | Given a `pattern` and a string `s`, find if `s` follows the same pattern.
Here **follow** means a full match, such that there is a bijection between a letter in `pattern` and a **non-empty** word in `s`.
**Example 1:**
**Input:** pattern = "abba ", s = "dog cat cat dog "
**Output:** true
**Example 2:**
**Input:*... | null |
Python 3 lines. Check one to one relationship | word-pattern | 0 | 1 | # Complexity\n- Time complexity:\n $$O(n)$$\n- Space complexity:\n $$O(n)$$\n\n# Code\n```\nclass Solution:\n def wordPattern(self, pattern: str, s: str) -> bool:\n ls = s.split()\n k, v, p = Counter(pattern), Counter(ls), Counter(zip(pattern, ls))\n return len(k) == len(v) == len(p) and l... | 5 | Given a `pattern` and a string `s`, find if `s` follows the same pattern.
Here **follow** means a full match, such that there is a bijection between a letter in `pattern` and a **non-empty** word in `s`.
**Example 1:**
**Input:** pattern = "abba ", s = "dog cat cat dog "
**Output:** true
**Example 2:**
**Input:*... | null |
✔️ [Python3] SINGLE HASHMAP (•̀ᴗ•́)و, Explained | word-pattern | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nAt the first glance, the problem can be solved simply by using a hashmap `w_to_p` which maps words to letters from the pattern. But consider this example: `w = [\'dog\', \'cat\']` and `p = \'aa\'`. In this case, the... | 166 | Given a `pattern` and a string `s`, find if `s` follows the same pattern.
Here **follow** means a full match, such that there is a bijection between a letter in `pattern` and a **non-empty** word in `s`.
**Example 1:**
**Input:** pattern = "abba ", s = "dog cat cat dog "
**Output:** true
**Example 2:**
**Input:*... | null |
Solution with dict | word-pattern | 0 | 1 | \n# Code\n```\nclass Solution:\n def wordPattern(self, pattern: str, s: str) -> bool:\n res = {}\n s = s.split()\n if len(pattern) != len(s):\n return False\n for i in range(len(pattern)):\n if pattern[i] not in res.keys():\n if s[i] not in res.values(... | 1 | Given a `pattern` and a string `s`, find if `s` follows the same pattern.
Here **follow** means a full match, such that there is a bijection between a letter in `pattern` and a **non-empty** word in `s`.
**Example 1:**
**Input:** pattern = "abba ", s = "dog cat cat dog "
**Output:** true
**Example 2:**
**Input:*... | null |
Python | Easy Solution✅ | word-pattern | 0 | 1 | ```\ndef wordPattern(self, pattern: str, s: str) -> bool:\n arr = s.split()\n if len(arr) != len(pattern):\n return False\n \n for i in range(len(arr)):\n if pattern.find(pattern[i]) != arr.index(arr[i]):\n return False\n return True\n``` | 9 | Given a `pattern` and a string `s`, find if `s` follows the same pattern.
Here **follow** means a full match, such that there is a bijection between a letter in `pattern` and a **non-empty** word in `s`.
**Example 1:**
**Input:** pattern = "abba ", s = "dog cat cat dog "
**Output:** true
**Example 2:**
**Input:*... | null |
2 liners python solution which beats 98% submissions | word-pattern | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo split the string and map it to the pattern based on its indes.\n\n# Complexity\n- Time complexity: O(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. $$O(n... | 2 | Given a `pattern` and a string `s`, find if `s` follows the same pattern.
Here **follow** means a full match, such that there is a bijection between a letter in `pattern` and a **non-empty** word in `s`.
**Example 1:**
**Input:** pattern = "abba ", s = "dog cat cat dog "
**Output:** true
**Example 2:**
**Input:*... | null |
python solution, beats 97%, 10 ms | word-pattern | 0 | 1 | # Code\n```\nclass Solution(object):\n def wordPattern(self, pattern, s):\n l = s.split()\n if len(pattern) != len(l):\n return False\n hash = {}\n for key in range(len(pattern)):\n if pattern[key] not in hash and l[key] in hash.values():\n return Fals... | 3 | Given a `pattern` and a string `s`, find if `s` follows the same pattern.
Here **follow** means a full match, such that there is a bijection between a letter in `pattern` and a **non-empty** word in `s`.
**Example 1:**
**Input:** pattern = "abba ", s = "dog cat cat dog "
**Output:** true
**Example 2:**
**Input:*... | null |
✅ EASIEST PYTHON ONE LINER SOLUTION ✅💯 | nim-game | 0 | 1 | \n# Code\n```\nclass Solution:\n def canWinNim(self, n: int) -> bool:\n return False if n % 4 == 0 else True\n \n\n``` | 2 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
PYTHON | Simple Solution | DP & Memorization & Math | nim-game | 0 | 1 | # Complexity\n- Time complexity: $$O(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```python\ndef can_win(n: int, cache: dict[int, bool]) -> bool:\n if n in cache:\n return cache[n]\n cache[n - 1] = can... | 2 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
Determining Winning Strategy in Nim Game using Game Theory. | nim-game | 0 | 1 | This is a classic example of a game theory problem where both players play optimally. The game\'s outcome depends on the number of stones in the heap. To determine whether you can win the game or not, we need to look at the number of stones in the heap and find a pattern.\n\nLet\'s consider the base cases first:\n\n- I... | 13 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
python || simple solution || one-liner | nim-game | 0 | 1 | **time complexity:** o(1)\n\n**space complexity:** o(1)\n\n```\nclass Solution:\n def canWinNim(self, n: int) -> bool:\n # if n is not a multiple of 4, it is possible to win\n return n % 4\n```\n\n\uD83D\uDD3C please upvote if helpful \uD83D\uDD3C | 5 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
🎮Nim Game || ⚡Beats 90%⚡ || 🫱🏻🫲🏼Beginners Friendly... | nim-game | 0 | 1 | # KARRAR\n>Math...\n>>Brainteaser...\n>>>Game theory...\n>>>>Nim game...\n>>>>>Optimzed and generalized...\n>>>>>>Beginners friendly...\n>>>>>>>Enjoy LeetCode...\n- PLEASE UPVOTE...\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n Take the mode of n by 4...\n ... | 1 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
292: Time 99.68% and Space 90.42%, Solution with step by step explanation | nim-game | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThis solution is similar to the previous one, but it is even more concise. Instead of using an if-else statement, it simply returns the result of the boolean expression n % 4 != 0. If n is divisible by 4, the expression will... | 7 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
Easy one step python solution with explanation | nim-game | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nbasically me loses if the num becomes 4 \nwhen n is a multiple of 4 there a n/4 number of 4\'s\nso irrespective of what me chooses the other person can choose a number to fill it up to 4 this goes on and the other person wins\nexample:\n4... | 4 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
Simplest Python code!! | nim-game | 0 | 1 | <!-- Describe your first thoughts on how to solve this problem. -->\n\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```... | 3 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
Python3 || 16ms, 99% faster|| one line | nim-game | 0 | 1 | ```\nclass Solution:\n def canWinNim(self, n: int) -> bool:\n return n % 4 != 0\n```\nif the value of n is divisible by 4 then only friend will win | 9 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
Bottom-up DP python | nim-game | 0 | 1 | Using bottom-up programming: \nWe define function F(x): The capability of you winning the game with x stones on your first turn.\nIf you go first and n <= 3: you win.\nelse F(x) > 3 after your turn, it\'s your opponents turn\n\nWhich means if you go first and each of your three possible moves all land on F(X) => True ... | 14 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
24ms 97% beats in Python3 one-line | nim-game | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n We "75%" always win if we take 3 stones\n# Approach\n<!-- Describe your approach to solving the problem. -->\n play this game and you will understand\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $... | 3 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
Python 1 line solution (BEATS 99.31%) | nim-game | 0 | 1 | # Intuition\nNotice that when we have a heap of 4 stones, we will lose. And similary, we will lose when we are left with a heap of 8 stones, because no matter you take 1,2, or 3 stones, your friend can always take 3, 2, or 1 stones. And now you are left with 4 stones. \n\n# Approach\nBased on the intuition, we simply c... | 0 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
Beats 100.00% of users || 0ms || Dynamic Programming 😍 | nim-game | 1 | 1 | ## Intuition\nThe problem involves determining whether you can win the game of Nim given a certain number of stones. The game of Nim is a two-player game where players take turns removing stones from piles. The key to winning Nim is to leave your opponent with a pile that has a number of stones that is a multiple of 4.... | 0 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
Getting to `canWinNim = lambda n: n % 4 != 0` | nim-game | 0 | 1 | # Intuition\n\nWe can win if one of our moves puts our opponent in a losing position. This recursive description gives us an initial solution:\n\n```python\nclass Solution:\n def canWinNim(self, n: int) -> bool:\n if n <= 3:\n return True\n else:\n return (\n not se... | 0 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
Explanation | nim-game | 0 | 1 | # Intuition\nWe find the pattern to win by listing first cases.\n\nLet denote **(n, r)** where **n** is the number of stones in the table and **r** be the result (win or lose) for the person who has a turn to remove these stones first.\n\n(1, win) (reason: remove 1 stone and win)\n(2, win) (remove 2 stones and win)\n(3... | 0 | You are playing the following Nim Game with your friend:
* Initially, there is a heap of stones on the table.
* You and your friend will alternate taking turns, and **you go first**.
* On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
* The one who removes the last stone is the... | If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner? |
logn | find-median-from-data-stream | 0 | 1 | \n\n# Complexity\n- Time complexity:\nO(logn)\n- Space complexity:\nO(N)\n# Code\n```\nclass MedianFinder:\n\n def __init__(self):\n self.small,self.large = [],[]\n \n\n def addNum(self, num: int) -> None:\n if self.large and num > self.large[0]:\n heapq.heappush(self.large,num)\n ... | 3 | The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
* For example, for `arr = [2,3,4]`, the median is `3`.
* For example, for `arr = [2,3]`, the median is `(2 + 3) / 2 = 2.5`.
Implement the M... | null |
Python || Min and Max Heap Solution | find-median-from-data-stream | 0 | 1 | ```\nclass MedianFinder:\n\n def __init__(self):\n self.maxHeap=[]\n self.minHeap=[]\n\n def addNum(self, num: int) -> None:\n n1=len(self.maxHeap)\n n2=len(self.minHeap)\n if n1==n2:\n if num>self.findMedian():\n heappush(self.minHeap,num)\n ... | 2 | The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
* For example, for `arr = [2,3,4]`, the median is `3`.
* For example, for `arr = [2,3]`, the median is `(2 + 3) / 2 = 2.5`.
Implement the M... | null |
using inbuilt functions easy | find-median-from-data-stream | 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)$$ --... | 5 | The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
* For example, for `arr = [2,3,4]`, the median is `3`.
* For example, for `arr = [2,3]`, the median is `(2 + 3) / 2 = 2.5`.
Implement the M... | null |
Using two heaps | find-median-from-data-stream | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\nBasically, you instantiate two heaps, a max heap to hold the first half of the numbers and a min_heap to hold the last half. Numbers will be added to the heaps in the sequence, max heap, min heap. When adding to the max_heap, you need to know if that ... | 2 | The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
* For example, for `arr = [2,3,4]`, the median is `3`.
* For example, for `arr = [2,3]`, the median is `(2 + 3) / 2 = 2.5`.
Implement the M... | null |
Simple 3 liner Python code | find-median-from-data-stream | 0 | 1 | SortedList makes life easier.\n\n# Code\n```\nfrom sortedcontainers import SortedList\nclass MedianFinder:\n\n def __init__(self):\n self.srt = SortedList()\n\n def addNum(self, num: int) -> None:\n self.srt.add(num)\n\n def findMedian(self) -> float:\n return median(self.srt)\n\n\n# Your ... | 3 | The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
* For example, for `arr = [2,3,4]`, the median is `3`.
* For example, for `arr = [2,3]`, the median is `(2 + 3) / 2 = 2.5`.
Implement the M... | null |
[Python] Logic Explained with 2 Heaps, Clean code. | find-median-from-data-stream | 0 | 1 | ```\n"""\n ## RC ##\n ## APPROACH : 2 HEAPS ##\n ## LOGIC ##\n ## One minheap to store low values and second maxheap to store max values, we keep track and update median every time after insertion ##\n \n\t\t## TIME COMPLEXITY : O(logN) ##\n\t\t## SPACE COMPLEXITY : O(N) ##\n\n ... | 79 | The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
* For example, for `arr = [2,3,4]`, the median is `3`.
* For example, for `arr = [2,3]`, the median is `(2 + 3) / 2 = 2.5`.
Implement the M... | null |
2 Heaps easy to understand solution | find-median-from-data-stream | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSince we need to find the median we are only concerened with the middle elements. We do not need to sort all the elements as we are worried about the elements other than the middle elements. THe question is how we can get the middle eleme... | 2 | The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
* For example, for `arr = [2,3,4]`, the median is `3`.
* For example, for `arr = [2,3]`, the median is `(2 + 3) / 2 = 2.5`.
Implement the M... | null |
295: Time 97.64%, Solution with step by step explanation | find-median-from-data-stream | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe MedianFinder class maintains two heaps: self.small (a max heap) and self.large (a min heap). The median is either the largest element in self.small, the smallest element in self.large, or the average of these two element... | 9 | The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
* For example, for `arr = [2,3,4]`, the median is `3`.
* For example, for `arr = [2,3]`, the median is `(2 + 3) / 2 = 2.5`.
Implement the M... | null |
Simple Solution with Approach using Python Heap | find-median-from-data-stream | 0 | 1 | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Create two heaps, such as mini and maxi.\n2. Mini would store smaller elements and maxi will store the larger elements following the max difference between their lengths should not exceed 1.\n3. In python, we only have min heap, in order to creat... | 2 | The **median** is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
* For example, for `arr = [2,3,4]`, the median is `3`.
* For example, for `arr = [2,3]`, the median is `(2 + 3) / 2 = 2.5`.
Implement the M... | null |
Recursive approach✅ | O( n)✅ | Python(Step by step explanation)✅ | serialize-and-deserialize-binary-tree | 0 | 1 | # Intuition\nThe problem involves serializing and deserializing a binary tree. Serialization is the process of converting a data structure into a string, and deserialization is the reverse process. The goal is to implement a `Codec` class that can serialize a binary tree into a string and deserialize the string back in... | 2 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
Easy || Understandable || Python Solution || Level Order Traversal || Striver || | serialize-and-deserialize-binary-tree | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\nUsed level order traversal\n\n\n# Code\n```\n# Definition for a binary tree node.\n# class TreeNode(object):\n# def __init__(self, x):\n# self.val = x\n# self.left = None\n# self.right = None\n\nclass Codec:\n\n def seri... | 1 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
✅ Explained - Simple and Clear Python3 Code✅ | serialize-and-deserialize-binary-tree | 0 | 1 | # Intuition\nThe Codec class provides methods for serializing and deserializing a binary tree. The intuition behind this code is that it encodes the structure of a binary tree into a string that can be easily transmitted or stored, and then decodes the string to reconstruct the original binary tree. This can be useful ... | 6 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
Iterative and Recursive (DFS/BFS) Approach easy understanding | serialize-and-deserialize-binary-tree | 0 | 1 | # Complexity\n- Time complexity: O(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# DFS\n```\nclass Codec:\n def serialize(self, root):\n if not root: return "#"\n return str(root.val) + "," + self.serialize... | 3 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
Bypass the code 😂 || Not recommended for learners || just 2 lines python | serialize-and-deserialize-binary-tree | 0 | 1 | ```\ngg = None\nclass Codec:\n def serialize(self, root):\n """Encodes a tree to a single string.\n \n :type root: TreeNode\n :rtype: str\n """\n global gg\n gg = root\n return \'\'\n \n\n def deserialize(self, data):\n """Decodes your encoded ... | 2 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
297: Time 95.90%, Solution with step by step explanation | serialize-and-deserialize-binary-tree | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThis code is an implementation of a binary tree serialization and deserialization algorithm.\n\nThe serialize method takes a binary tree root node as input and returns a string representation of the tree. It uses a pre-order... | 6 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
Python || Easy || BFS || Queue | serialize-and-deserialize-binary-tree | 0 | 1 | ```\nclass Codec:\n\n def serialize(self, root):\n if root==None:\n return ""\n q=deque([root])\n s=""\n while q:\n node=q.popleft()\n if node:\n s+=str(node.val)+" "\n q.append(node.left)\n q.append(node.right)... | 2 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
[Python] BFS to serialize like LeetCode | serialize-and-deserialize-binary-tree | 0 | 1 | ```\nclass Codec:\n\n def serialize(self, root):\n # use level order traversal to match LeetCode\'s serialization format\n flat_bt = []\n queue = collections.deque([root])\n while queue:\n node = queue.pop()\n if node:\n flat_bt.append(str(node.val))\n... | 23 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
Easy to understand recursive in Python3 | serialize-and-deserialize-binary-tree | 0 | 1 | \n```\nclass Codec:\n\n def serialize(self, root):\n if not root:\n return ""\n\n return f"{root.val},{self.serialize(root.left)},{self.serialize(root.right)}"\n\n\n def deserialize(self, data: str):\n if not data:\n return None\n\n return self.deserialize_list(da... | 16 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
[Python3] DFS Easy to Understand Solution | serialize-and-deserialize-binary-tree | 0 | 1 | ```\n\nclass Codec:\n\n def serialize(self, root):\n res = []\n def preorder(root):\n if not root:\n res.append(\'N\')\n return\n res.append(str(root.val))\n preorder(root.left)\n preorder(root.right)\n preorder(root)\n ... | 5 | Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a bi... | null |
Easy Python Solution Using Counter | bulls-and-cows | 0 | 1 | # Code\n```\nclass Solution:\n def getHint(self, secret: str, guess: str) -> str:\n dic=Counter(secret)-Counter(guess)\n cnt=0\n for i in range(len(secret)):\n if secret[i]==guess[i]:\n cnt+=1\n cnt2=len(secret)-sum(dic.values())-cnt\n return str(cnt)+"A"+... | 4 | You are playing the **[Bulls and Cows](https://en.wikipedia.org/wiki/Bulls_and_Cows)** game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
* The number of "bulls ", which are digits in the ... | null |
✔Easy Python solution using Hashmap | bulls-and-cows | 0 | 1 | # Intuition\nFor bulls we have to check for the digits having same position in both the strings i.e. secret and guess, then will count for the rest of the digits which are common in both strings.\n\n# Approach\nWe will take **two dictionary** for **counting the frequency** of each digit occured in the respective string... | 4 | You are playing the **[Bulls and Cows](https://en.wikipedia.org/wiki/Bulls_and_Cows)** game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
* The number of "bulls ", which are digits in the ... | null |
python3, simple solution, defaultdict() for cows | bulls-and-cows | 0 | 1 | ERROR: type should be string, got "https://leetcode.com/submissions/detail/875564006/\\nRuntime: **30 ms**, faster than 99.56% of Python3 online submissions for Bulls and Cows. \\nMemory Usage: 14 MB, less than 29.43% of Python3 online submissions for Bulls and Cows. \\n```\\nclass Solution:\\n def getHint(self, secret: str, guess: str) -> str:\\n bulls, cows, ds, dg = 0, 0, defaultdict(lambda:0), defaultdict(lambda:0)\\n for s,g in zip(secret, guess):\\n if s==g:\\n bulls += 1\\n continue\\n ds[s]+=1; dg[g]+=1\\n for s in ds:\\n if s in dg:\\n cows += min(ds[s], dg[s])\\n return f\\'{bulls}A{cows}B\\'\\n```" | 3 | You are playing the **[Bulls and Cows](https://en.wikipedia.org/wiki/Bulls_and_Cows)** game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
* The number of "bulls ", which are digits in the ... | null |
Python Easy Solution (Beats 99.5%) | bulls-and-cows | 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)$$ --... | 3 | You are playing the **[Bulls and Cows](https://en.wikipedia.org/wiki/Bulls_and_Cows)** game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
* The number of "bulls ", which are digits in the ... | null |
299: Time 98.43% and Space 98.71%, Solution with step by step explanation | bulls-and-cows | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Define the function getHint that takes two strings secret and guess as input and returns a string.\n2. Initialize two dictionaries secret_dict and guess_dict to store the frequencies of digits in the secret and guess stri... | 6 | You are playing the **[Bulls and Cows](https://en.wikipedia.org/wiki/Bulls_and_Cows)** game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
* The number of "bulls ", which are digits in the ... | null |
✔️ Python O(n) Solution Using Hashmap with Explanation | 99% Faster 🔥 | bulls-and-cows | 0 | 1 | **\uD83D\uDD3C IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE \uD83D\uDC4D**\n\n```\nclass Solution:\n def getHint(self, secret: str, guess: str) -> str:\n # Dictionary for Lookup\n lookup = Counter(secret)\n \n x, y = 0, 0\n \n # First finding numbers which are at correct posi... | 8 | You are playing the **[Bulls and Cows](https://en.wikipedia.org/wiki/Bulls_and_Cows)** game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
* The number of "bulls ", which are digits in the ... | null |
Fast and easy to understand Python solution O(n) | bulls-and-cows | 0 | 1 | ```\nclass Solution:\n def getHint(self, secret: str, guess: str) -> str:\n \n\t\t# The main idea is to understand that cow cases contain the bull cases\n\t\t\n\t\t# This loop will take care of "bull" cases\n bull=0\n for i in range(len(secret)):\n bull += int(secret[i] == guess[i])\n... | 20 | You are playing the **[Bulls and Cows](https://en.wikipedia.org/wiki/Bulls_and_Cows)** game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
* The number of "bulls ", which are digits in the ... | null |
[Python 3] Hashmap Solution. Easy to understand. | bulls-and-cows | 0 | 1 | **Python3:**\n```\nclass Solution:\n def getHint(self, secret: str, guess: str) -> str:\n nums = collections.Counter(secret) # Counting occurences of each digit in secret.\n cows = 0\n bulls = 0\n \n # Counting Cows\n for n in guess:\n if n in nums:\n ... | 1 | You are playing the **[Bulls and Cows](https://en.wikipedia.org/wiki/Bulls_and_Cows)** game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
* The number of "bulls ", which are digits in the ... | null |
2-pass & 1-pass solution Python | bulls-and-cows | 0 | 1 | **2-pass solution**\n```\nfrom collections import Counter\n\nclass Solution:\n def getHint(self, secret: str, guess: str) -> str:\n secret_non_bull, guess_non_bull = [], []\n num_bulls = 0\n for char1, char2 in zip(secret, guess):\n if char1 == char2:\n num_bulls += 1\n... | 4 | You are playing the **[Bulls and Cows](https://en.wikipedia.org/wiki/Bulls_and_Cows)** game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
* The number of "bulls ", which are digits in the ... | null |
Python3 || 7 lines, binSearch, cheating, w/explanation || T/M: 94%/82% | longest-increasing-subsequence | 0 | 1 | ```\nclass Solution: # Suppose, for example:\n # nums = [1,8,4,5,3,7],\n # for which the longest strictly increasing subsequence is arr = [1,4,5,7],\n # giving len(arr) = 4 as the answer\n #\n # Here\'s the plan... | 72 | Given an integer array `nums`, return _the length of the longest **strictly increasing**_ _**subsequence**_.
**Example 1:**
**Input:** nums = \[10,9,2,5,3,7,101,18\]
**Output:** 4
**Explanation:** The longest increasing subsequence is \[2,3,7,101\], therefore the length is 4.
**Example 2:**
**Input:** nums = \[0,1,... | null |
🔥 Python || Fast Than 95% ✅ || Less Than 83% ✅ || O(nlogn) | longest-increasing-subsequence | 0 | 1 | **Appreciate if you could upvote this solution**\n\n**Method**: `DP`\n\n```\ndef lengthOfLIS(self, nums: List[int]) -> int:\n\ttotal_number = len(nums)\n\tdp = [0 for _ in range(total_number)]\n\tfor i in range(1, total_number):\n\t\tfor j in range(i):\n\t\t\tif nums[j] < nums[i]:\n\t\t\t\tdp[i] = max(dp[i], dp[j] + 1)... | 46 | Given an integer array `nums`, return _the length of the longest **strictly increasing**_ _**subsequence**_.
**Example 1:**
**Input:** nums = \[10,9,2,5,3,7,101,18\]
**Output:** 4
**Explanation:** The longest increasing subsequence is \[2,3,7,101\], therefore the length is 4.
**Example 2:**
**Input:** nums = \[0,1,... | null |
✅ PYTHON || Beginner Friendly || O(n^2) 🔥🔥🔥 | longest-increasing-subsequence | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe want to find the length of the Longest Increasing Subsequence (LIS) in the given list of numbers.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. We will use dynamic programming to solve this problem. We\'ll cr... | 3 | Given an integer array `nums`, return _the length of the longest **strictly increasing**_ _**subsequence**_.
**Example 1:**
**Input:** nums = \[10,9,2,5,3,7,101,18\]
**Output:** 4
**Explanation:** The longest increasing subsequence is \[2,3,7,101\], therefore the length is 4.
**Example 2:**
**Input:** nums = \[0,1,... | null |
Binbin's another brilliant/wise idea! | longest-increasing-subsequence | 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. $$O(n)$$ --... | 2 | Given an integer array `nums`, return _the length of the longest **strictly increasing**_ _**subsequence**_.
**Example 1:**
**Input:** nums = \[10,9,2,5,3,7,101,18\]
**Output:** 4
**Explanation:** The longest increasing subsequence is \[2,3,7,101\], therefore the length is 4.
**Example 2:**
**Input:** nums = \[0,1,... | null |
backtracking approach | remove-invalid-parentheses | 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. $$O(n)$$ --... | 2 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
backtracking approach | remove-invalid-parentheses | 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. $$O(n)$$ --... | 2 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Using 2 windows and deleting from forward and reversed string | remove-invalid-parentheses | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n print("Running for ", s)\n print\n if not s: \n return [""]\n\n ans = []\n\n\n def remove(updated, start, remStart, pairs = ["(", ")"]):\n # print("S:", s, "start: ",... | 1 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
Using 2 windows and deleting from forward and reversed string | remove-invalid-parentheses | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n print("Running for ", s)\n print\n if not s: \n return [""]\n\n ans = []\n\n\n def remove(updated, start, remStart, pairs = ["(", ")"]):\n # print("S:", s, "start: ",... | 1 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
[Python] Backtracking solution, detailed explanation | remove-invalid-parentheses | 0 | 1 | ```\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n ## RC ##\n ## APPROACH : BACK-TRACKING ##\n ## Similar to Leetcode 32. Longest Valid Parentheses ##\n ## LOGIC ##\n # 1. use stack to find invalid left and right braces.\n # 2. if its close... | 25 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
[Python] Backtracking solution, detailed explanation | remove-invalid-parentheses | 0 | 1 | ```\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n ## RC ##\n ## APPROACH : BACK-TRACKING ##\n ## Similar to Leetcode 32. Longest Valid Parentheses ##\n ## LOGIC ##\n # 1. use stack to find invalid left and right braces.\n # 2. if its close... | 25 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
301: Time 90.95%, Solution with step by step explanation | remove-invalid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n\n1. getLeftAndRightCounts(s: str) -> tuple: This function takes a string as input and returns a tuple containing two integers. The two integers are the counts of the left and right parentheses that need to be removed to mak... | 5 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
301: Time 90.95%, Solution with step by step explanation | remove-invalid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n\n1. getLeftAndRightCounts(s: str) -> tuple: This function takes a string as input and returns a tuple containing two integers. The two integers are the counts of the left and right parentheses that need to be removed to mak... | 5 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Easy to understand Python 🐍 solution | remove-invalid-parentheses | 0 | 1 | ```\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n \n def valid(s):\n l,r=0,0\n for c in s:\n if c==\'(\':\n l+=1\n elif c==\')\':\n if l<=0:\n r+=1\n ... | 3 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
Easy to understand Python 🐍 solution | remove-invalid-parentheses | 0 | 1 | ```\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n \n def valid(s):\n l,r=0,0\n for c in s:\n if c==\'(\':\n l+=1\n elif c==\')\':\n if l<=0:\n r+=1\n ... | 3 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Python 3 commented BFS | remove-invalid-parentheses | 0 | 1 | *Runtime: 196 ms, faster than 45.23% of Python3 online submissions for Remove Invalid Parentheses.\nMemory Usage: 14.3 MB, less than 26.77% of Python3 online submissions for Remove Invalid Parentheses.*\n\n```python\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n # define when a ... | 9 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
Python 3 commented BFS | remove-invalid-parentheses | 0 | 1 | *Runtime: 196 ms, faster than 45.23% of Python3 online submissions for Remove Invalid Parentheses.\nMemory Usage: 14.3 MB, less than 26.77% of Python3 online submissions for Remove Invalid Parentheses.*\n\n```python\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n # define when a ... | 9 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Python solution using Smart Backtracking with a Simple Trick Beats 82% in Time and 100% in Space | remove-invalid-parentheses | 0 | 1 | This solution uses backtracking but in order to prevent repeatitive calls of the backtracking function, I first calculate how many left and right parenthesis needs to be removed. If a right parenthesis doesn\'t have a matching left one, it needs to be removed, also if there are more left parenthesis compared to the rig... | 11 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
Python solution using Smart Backtracking with a Simple Trick Beats 82% in Time and 100% in Space | remove-invalid-parentheses | 0 | 1 | This solution uses backtracking but in order to prevent repeatitive calls of the backtracking function, I first calculate how many left and right parenthesis needs to be removed. If a right parenthesis doesn\'t have a matching left one, it needs to be removed, also if there are more left parenthesis compared to the rig... | 11 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Simple backtracking logic | remove-invalid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nBacktracking\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFind min open and close braces to remove and then run dfs\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(2... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
Simple backtracking logic | remove-invalid-parentheses | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nBacktracking\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFind min open and close braces to remove and then run dfs\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(2... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Simpler than official solution | remove-invalid-parentheses | 0 | 1 | # Intuition\nWe\'d need to formulate every combination and see which ones are valid. A combinatorial problem.\n\n# Approach\nUse backtracking and collect only valid paranthesis\n\n# Complexity\n- Time complexity: Since every paranthesis can either be included or not => 2 options => O(2 ^ N)\n\n- Space complexity: curre... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
Simpler than official solution | remove-invalid-parentheses | 0 | 1 | # Intuition\nWe\'d need to formulate every combination and see which ones are valid. A combinatorial problem.\n\n# Approach\nUse backtracking and collect only valid paranthesis\n\n# Complexity\n- Time complexity: Since every paranthesis can either be included or not => 2 options => O(2 ^ N)\n\n- Space complexity: curre... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Python, beats 80% slightly different route | remove-invalid-parentheses | 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 string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
Python, beats 80% slightly different route | remove-invalid-parentheses | 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 string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Concise solution on python3 | remove-invalid-parentheses | 0 | 1 | \n# Code\n```\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n\n n = len(s)\n mx = 0\n ans = set()\n \n def go(pos, res, bal):\n nonlocal ans, mx\n\n if bal > 0:\n return\n\n if pos == n:\n ... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
Concise solution on python3 | remove-invalid-parentheses | 0 | 1 | \n# Code\n```\nclass Solution:\n def removeInvalidParentheses(self, s: str) -> List[str]:\n\n n = len(s)\n mx = 0\n ans = set()\n \n def go(pos, res, bal):\n nonlocal ans, mx\n\n if bal > 0:\n return\n\n if pos == n:\n ... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
python3 solution | remove-invalid-parentheses | 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 string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
python3 solution | remove-invalid-parentheses | 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 string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Python beats 99% solution | remove-invalid-parentheses | 0 | 1 | Not sure if it\'s because this question is too old to have enough Python submissions (very likely), and not sure if there are already similar or the same solutions. but I tried to digest the Editorial thoughts and convert it to code as concise/clear as possible.\n# Intuition\n<!-- Describe your first thoughts on how to... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
Python beats 99% solution | remove-invalid-parentheses | 0 | 1 | Not sure if it\'s because this question is too old to have enough Python submissions (very likely), and not sure if there are already similar or the same solutions. but I tried to digest the Editorial thoughts and convert it to code as concise/clear as possible.\n# Intuition\n<!-- Describe your first thoughts on how to... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
It ain't much but it's honest work. [Python][stack][DFS] | remove-invalid-parentheses | 0 | 1 | # Intuition\n- Break this down into two problems\n\n# Approach\nFor me, there are two parts to this problem\n- Find k: minimum number of invalid parentheses to remove [LC 1249](https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/description/)\n- Once you have k, it\'s a simple DFS: `dfs(index, k, pat... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
It ain't much but it's honest work. [Python][stack][DFS] | remove-invalid-parentheses | 0 | 1 | # Intuition\n- Break this down into two problems\n\n# Approach\nFor me, there are two parts to this problem\n- Find k: minimum number of invalid parentheses to remove [LC 1249](https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/description/)\n- Once you have k, it\'s a simple DFS: `dfs(index, k, pat... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
So many parameters | remove-invalid-parentheses | 0 | 1 | # Intuition\nWe only want to remove as many opening or closing parentheses as possible, to this end, we first find out how many of each we\'d need to remove, then run a recursive search to find all possible valid expressions.\n\n# Approach\n - First, find the number of closing or opening parentheses to be removed\n ... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.... |
So many parameters | remove-invalid-parentheses | 0 | 1 | # Intuition\nWe only want to remove as many opening or closing parentheses as possible, to this end, we first find out how many of each we\'d need to remove, then run a recursive search to find all possible valid expressions.\n\n# Approach\n - First, find the number of closing or opening parentheses to be removed\n ... | 0 | Given a string `s` that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return _a list of **unique strings** that are valid with the minimum number of removals_. You may return the answer in **any order**.
**Example 1:**
**Input:** s = "()())() "
**... | Since we don't know which of the brackets can possibly be removed, we try out all the options! We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:
We keep the bracket and add it to the expression that we are building on the fly during recursion.
O... |
Python O(1) solution, beats 97.85% | range-sum-query-immutable | 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$$O(1)$$\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 3 | Given an integer array `nums`, handle multiple queries of the following type:
1. Calculate the **sum** of the elements of `nums` between indices `left` and `right` **inclusive** where `left <= right`.
Implement the `NumArray` class:
* `NumArray(int[] nums)` Initializes the object with the integer array `nums`.
* ... | null |
Easy to Understand | Faster | Simple | Python Solution | range-sum-query-immutable | 0 | 1 | ```\n def __init__(self, nums: List[int]):\n self.sum = []\n sum_till = 0\n for i in nums:\n sum_till += i\n self.sum.append(sum_till)\n\n def sumRange(self, i: int, j: int) -> int:\n if i > 0 and j > 0:\n return self.sum[j] - self.sum[i - 1]\n e... | 43 | Given an integer array `nums`, handle multiple queries of the following type:
1. Calculate the **sum** of the elements of `nums` between indices `left` and `right` **inclusive** where `left <= right`.
Implement the `NumArray` class:
* `NumArray(int[] nums)` Initializes the object with the integer array `nums`.
* ... | null |
PYTHON3 SINGLE LINE CODE ✅✅✅ | range-sum-query-immutable | 0 | 1 | \n\n# Code\n```\nclass NumArray:\n\n def __init__(self, nums: List[int]):\n self.nums=nums\n\n def sumRange(self, left: int, right: int) -> int:\n return sum(self.nums[left:right+1])\n\n\n# Your NumArray object will be instantiated and called as such:\n# obj = NumArray(nums)\n# param_1 = obj.sumRang... | 1 | Given an integer array `nums`, handle multiple queries of the following type:
1. Calculate the **sum** of the elements of `nums` between indices `left` and `right` **inclusive** where `left <= right`.
Implement the `NumArray` class:
* `NumArray(int[] nums)` Initializes the object with the integer array `nums`.
* ... | null |
303: Solution with step by step explanation | range-sum-query-immutable | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n- We define a NumArray class with two methods: __init__ and sumRange.\n- In __init__, we compute the prefix sums of the input nums array and store them in a prefix_sum list. We initialize the prefix_sum list with 0\'s and th... | 7 | Given an integer array `nums`, handle multiple queries of the following type:
1. Calculate the **sum** of the elements of `nums` between indices `left` and `right` **inclusive** where `left <= right`.
Implement the `NumArray` class:
* `NumArray(int[] nums)` Initializes the object with the integer array `nums`.
* ... | null |
Efficient Range Sum Query using Prefix Sum Array | range-sum-query-immutable | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to find the sum of the elements of nums between indices left and right inclusive. This can be achieved by creating a prefix sum array, which stores the sum of the elements up to that index. By subtracting the prefi... | 5 | Given an integer array `nums`, handle multiple queries of the following type:
1. Calculate the **sum** of the elements of `nums` between indices `left` and `right` **inclusive** where `left <= right`.
Implement the `NumArray` class:
* `NumArray(int[] nums)` Initializes the object with the integer array `nums`.
* ... | null |
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