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 Math | number-of-burgers-with-no-waste-of-ingredients | 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\nO(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 0 | You are given a circle represented as `(radius, xCenter, yCenter)` and an axis-aligned rectangle represented as `(x1, y1, x2, y2)`, where `(x1, y1)` are the coordinates of the bottom-left corner, and `(x2, y2)` are the coordinates of the top-right corner of the rectangle.
Return `true` _if the circle and rectangle are... | Can we have an answer if the number of tomatoes is odd ? If we have answer will be there multiple answers or just one answer ? Let us define number of jumbo burgers as X and number of small burgers as Y
We have to find an x and y in this equation 1. 4X + 2Y = tomato 2. X + Y = cheese |
Python 3 | Solution | one-liner | number-of-burgers-with-no-waste-of-ingredients | 0 | 1 | # Code\n```\nclass Solution:\n def numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:\n\n return [tomatoSlices//2 - cheeseSlices, 2*cheeseSlices - tomatoSlices//2] if (tomatoSlices - 2*cheeseSlices)%2 == 0 and (tomatoSlices - 2*cheeseSlices) >= 0 and (2*cheeseSlices - tomatoSlices//2) >= 0... | 0 | Given two integers `tomatoSlices` and `cheeseSlices`. The ingredients of different burgers are as follows:
* **Jumbo Burger:** `4` tomato slices and `1` cheese slice.
* **Small Burger:** `2` Tomato slices and `1` cheese slice.
Return `[total_jumbo, total_small]` so that the number of remaining `tomatoSlices` equa... | Find the divisors of n+1 and n+2. To find the divisors of a number, you only need to iterate to the square root of that number. |
Python 3 | Solution | one-liner | number-of-burgers-with-no-waste-of-ingredients | 0 | 1 | # Code\n```\nclass Solution:\n def numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:\n\n return [tomatoSlices//2 - cheeseSlices, 2*cheeseSlices - tomatoSlices//2] if (tomatoSlices - 2*cheeseSlices)%2 == 0 and (tomatoSlices - 2*cheeseSlices) >= 0 and (2*cheeseSlices - tomatoSlices//2) >= 0... | 0 | You are given a circle represented as `(radius, xCenter, yCenter)` and an axis-aligned rectangle represented as `(x1, y1, x2, y2)`, where `(x1, y1)` are the coordinates of the bottom-left corner, and `(x2, y2)` are the coordinates of the top-right corner of the rectangle.
Return `true` _if the circle and rectangle are... | Can we have an answer if the number of tomatoes is odd ? If we have answer will be there multiple answers or just one answer ? Let us define number of jumbo burgers as X and number of small burgers as Y
We have to find an x and y in this equation 1. 4X + 2Y = tomato 2. X + Y = cheese |
Python exercise in solving linear equations | number-of-burgers-with-no-waste-of-ingredients | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLetting T = ```tomatoSlices``` and C = ```cheeseSlices```, we have the equations:\n T = tomatoSlices; C = cheeseSlices\n T - ( 4*J + 2*S ) = 0\n C - ( 1*J + 1*S ) = 0\n T = 4*J + 2*S\n C = 1*J + 1*S\n ... | 0 | Given two integers `tomatoSlices` and `cheeseSlices`. The ingredients of different burgers are as follows:
* **Jumbo Burger:** `4` tomato slices and `1` cheese slice.
* **Small Burger:** `2` Tomato slices and `1` cheese slice.
Return `[total_jumbo, total_small]` so that the number of remaining `tomatoSlices` equa... | Find the divisors of n+1 and n+2. To find the divisors of a number, you only need to iterate to the square root of that number. |
Python exercise in solving linear equations | number-of-burgers-with-no-waste-of-ingredients | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nLetting T = ```tomatoSlices``` and C = ```cheeseSlices```, we have the equations:\n T = tomatoSlices; C = cheeseSlices\n T - ( 4*J + 2*S ) = 0\n C - ( 1*J + 1*S ) = 0\n T = 4*J + 2*S\n C = 1*J + 1*S\n ... | 0 | You are given a circle represented as `(radius, xCenter, yCenter)` and an axis-aligned rectangle represented as `(x1, y1, x2, y2)`, where `(x1, y1)` are the coordinates of the bottom-left corner, and `(x2, y2)` are the coordinates of the top-right corner of the rectangle.
Return `true` _if the circle and rectangle are... | Can we have an answer if the number of tomatoes is odd ? If we have answer will be there multiple answers or just one answer ? Let us define number of jumbo burgers as X and number of small burgers as Y
We have to find an x and y in this equation 1. 4X + 2Y = tomato 2. X + Y = cheese |
mind blown logic using DP in python3 | count-square-submatrices-with-all-ones | 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 `m * n` matrix of ones and zeros, return how many **square** submatrices have all ones.
**Example 1:**
**Input:** matrix =
\[
\[0,1,1,1\],
\[1,1,1,1\],
\[0,1,1,1\]
\]
**Output:** 15
**Explanation:**
There are **10** squares of side 1.
There are **4** squares of side 2.
There is **1** square of side 3.... | A number is a multiple of three if and only if its sum of digits is a multiple of three. Use dynamic programming. To find the maximum number, try to maximize the number of digits of the number. Sort the digits in descending order to find the maximum number. |
mind blown logic using DP in python3 | count-square-submatrices-with-all-ones | 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 | A chef has collected data on the `satisfaction` level of his `n` dishes. Chef can cook any dish in 1 unit of time.
**Like-time coefficient** of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. `time[i] * satisfaction[i]`.
Return _the maximum sum... | Create an additive table that counts the sum of elements of submatrix with the superior corner at (0,0). Loop over all subsquares in O(n^3) and check if the sum make the whole array to be ones, if it checks then add 1 to the answer. |
DP with figure explanation | count-square-submatrices-with-all-ones | 0 | 1 | \n\n----\n----\n## ## Explanation\n\n `dp[i][j]` represent edge length of the biggest square whose lower right corner element is `matrix[i][j]`. Also there are `dp[i][j]` squares whose lower right corner element are `matrix[i][j]`. The answer of c... | 154 | Given a `m * n` matrix of ones and zeros, return how many **square** submatrices have all ones.
**Example 1:**
**Input:** matrix =
\[
\[0,1,1,1\],
\[1,1,1,1\],
\[0,1,1,1\]
\]
**Output:** 15
**Explanation:**
There are **10** squares of side 1.
There are **4** squares of side 2.
There is **1** square of side 3.... | A number is a multiple of three if and only if its sum of digits is a multiple of three. Use dynamic programming. To find the maximum number, try to maximize the number of digits of the number. Sort the digits in descending order to find the maximum number. |
DP with figure explanation | count-square-submatrices-with-all-ones | 0 | 1 | \n\n----\n----\n## ## Explanation\n\n `dp[i][j]` represent edge length of the biggest square whose lower right corner element is `matrix[i][j]`. Also there are `dp[i][j]` squares whose lower right corner element are `matrix[i][j]`. The answer of c... | 154 | A chef has collected data on the `satisfaction` level of his `n` dishes. Chef can cook any dish in 1 unit of time.
**Like-time coefficient** of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. `time[i] * satisfaction[i]`.
Return _the maximum sum... | Create an additive table that counts the sum of elements of submatrix with the superior corner at (0,0). Loop over all subsquares in O(n^3) and check if the sum make the whole array to be ones, if it checks then add 1 to the answer. |
Superb Logic Python3 ---->DP | count-square-submatrices-with-all-ones | 0 | 1 | # Dynamic Programming\n```\nclass Solution:\n def countSquares(self, matrix: List[List[int]]) -> int:\n row=len(matrix)\n col=len(matrix[0])\n dp=[[0]*col for i in range(row)]\n for i in range(row):\n for j in range(col):\n if (i==0 or j==0) and matrix[i][j]==1:\... | 7 | Given a `m * n` matrix of ones and zeros, return how many **square** submatrices have all ones.
**Example 1:**
**Input:** matrix =
\[
\[0,1,1,1\],
\[1,1,1,1\],
\[0,1,1,1\]
\]
**Output:** 15
**Explanation:**
There are **10** squares of side 1.
There are **4** squares of side 2.
There is **1** square of side 3.... | A number is a multiple of three if and only if its sum of digits is a multiple of three. Use dynamic programming. To find the maximum number, try to maximize the number of digits of the number. Sort the digits in descending order to find the maximum number. |
Superb Logic Python3 ---->DP | count-square-submatrices-with-all-ones | 0 | 1 | # Dynamic Programming\n```\nclass Solution:\n def countSquares(self, matrix: List[List[int]]) -> int:\n row=len(matrix)\n col=len(matrix[0])\n dp=[[0]*col for i in range(row)]\n for i in range(row):\n for j in range(col):\n if (i==0 or j==0) and matrix[i][j]==1:\... | 7 | A chef has collected data on the `satisfaction` level of his `n` dishes. Chef can cook any dish in 1 unit of time.
**Like-time coefficient** of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. `time[i] * satisfaction[i]`.
Return _the maximum sum... | Create an additive table that counts the sum of elements of submatrix with the superior corner at (0,0). Loop over all subsquares in O(n^3) and check if the sum make the whole array to be ones, if it checks then add 1 to the answer. |
Py | Easy solution | Best Approch 💯 ✅ | count-square-submatrices-with-all-ones | 0 | 1 | ```\nclass Solution:\n def countSquares(self, arr: List[List[int]]) -> int:\n n=len(arr)\n m=len(arr[0])\n dp=[[0 for i in range(m)]for j in range(n)]\n for i in range(n):\n dp[i][0]=arr[i][0]\n for j in range(m):\n dp[0][j]=arr[0][j]\n for i in range(1... | 1 | Given a `m * n` matrix of ones and zeros, return how many **square** submatrices have all ones.
**Example 1:**
**Input:** matrix =
\[
\[0,1,1,1\],
\[1,1,1,1\],
\[0,1,1,1\]
\]
**Output:** 15
**Explanation:**
There are **10** squares of side 1.
There are **4** squares of side 2.
There is **1** square of side 3.... | A number is a multiple of three if and only if its sum of digits is a multiple of three. Use dynamic programming. To find the maximum number, try to maximize the number of digits of the number. Sort the digits in descending order to find the maximum number. |
Py | Easy solution | Best Approch 💯 ✅ | count-square-submatrices-with-all-ones | 0 | 1 | ```\nclass Solution:\n def countSquares(self, arr: List[List[int]]) -> int:\n n=len(arr)\n m=len(arr[0])\n dp=[[0 for i in range(m)]for j in range(n)]\n for i in range(n):\n dp[i][0]=arr[i][0]\n for j in range(m):\n dp[0][j]=arr[0][j]\n for i in range(1... | 1 | A chef has collected data on the `satisfaction` level of his `n` dishes. Chef can cook any dish in 1 unit of time.
**Like-time coefficient** of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. `time[i] * satisfaction[i]`.
Return _the maximum sum... | Create an additive table that counts the sum of elements of submatrix with the superior corner at (0,0). Loop over all subsquares in O(n^3) and check if the sum make the whole array to be ones, if it checks then add 1 to the answer. |
Python || DP || Tabulation || Explained | count-square-submatrices-with-all-ones | 0 | 1 | **Intuition:**\n\nFollowing the tabulation method, we will first create a 2D dp array of the same size as the given 2D matrix. And in the dp array, dp[i][j] will signify, how many squares end at the cell (i, j) i.e. for how many squares the rightmost bottom cell is (i, j).\nFor example, consider the following matrix:\n... | 2 | Given a `m * n` matrix of ones and zeros, return how many **square** submatrices have all ones.
**Example 1:**
**Input:** matrix =
\[
\[0,1,1,1\],
\[1,1,1,1\],
\[0,1,1,1\]
\]
**Output:** 15
**Explanation:**
There are **10** squares of side 1.
There are **4** squares of side 2.
There is **1** square of side 3.... | A number is a multiple of three if and only if its sum of digits is a multiple of three. Use dynamic programming. To find the maximum number, try to maximize the number of digits of the number. Sort the digits in descending order to find the maximum number. |
Python || DP || Tabulation || Explained | count-square-submatrices-with-all-ones | 0 | 1 | **Intuition:**\n\nFollowing the tabulation method, we will first create a 2D dp array of the same size as the given 2D matrix. And in the dp array, dp[i][j] will signify, how many squares end at the cell (i, j) i.e. for how many squares the rightmost bottom cell is (i, j).\nFor example, consider the following matrix:\n... | 2 | A chef has collected data on the `satisfaction` level of his `n` dishes. Chef can cook any dish in 1 unit of time.
**Like-time coefficient** of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. `time[i] * satisfaction[i]`.
Return _the maximum sum... | Create an additive table that counts the sum of elements of submatrix with the superior corner at (0,0). Loop over all subsquares in O(n^3) and check if the sum make the whole array to be ones, if it checks then add 1 to the answer. |
Easy DP Solution | palindrome-partitioning-iii | 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 | You are given a string `s` containing lowercase letters and an integer `k`. You need to :
* First, change some characters of `s` to other lowercase English letters.
* Then divide `s` into `k` non-empty disjoint substrings such that each substring is a palindrome.
Return _the minimal number of characters that you ... | null |
Easy DP Solution | palindrome-partitioning-iii | 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 `nums`, obtain a subsequence of the array whose sum of elements is **strictly greater** than the sum of the non included elements in such subsequence.
If there are multiple solutions, return the subsequence with **minimum size** and if there still exist multiple solutions, return the subsequence with t... | For each substring calculate the minimum number of steps to make it palindrome and store it in a table. Create a dp(pos, cnt) which means the minimum number of characters changed for the suffix of s starting on pos splitting the suffix on cnt chunks. |
python super easy DP top down | palindrome-partitioning-iii | 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 | You are given a string `s` containing lowercase letters and an integer `k`. You need to :
* First, change some characters of `s` to other lowercase English letters.
* Then divide `s` into `k` non-empty disjoint substrings such that each substring is a palindrome.
Return _the minimal number of characters that you ... | null |
python super easy DP top down | palindrome-partitioning-iii | 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 `nums`, obtain a subsequence of the array whose sum of elements is **strictly greater** than the sum of the non included elements in such subsequence.
If there are multiple solutions, return the subsequence with **minimum size** and if there still exist multiple solutions, return the subsequence with t... | For each substring calculate the minimum number of steps to make it palindrome and store it in a table. Create a dp(pos, cnt) which means the minimum number of characters changed for the suffix of s starting on pos splitting the suffix on cnt chunks. |
Pandas vs SQL | Elegant & Short | All 30 Days of Pandas solutions ✅ | students-and-examinations | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```Python []\ndef students_and_examinations(students: pd.DataFrame, subjects: pd.DataFrame, examinations: pd.DataFrame) -> pd.DataFrame:\n return pd.merge(\n left=pd.merge(\n students, subjects, how=\'cross\',\n ... | 38 | A **happy string** is a string that:
* consists only of letters of the set `['a', 'b', 'c']`.
* `s[i] != s[i + 1]` for all values of `i` from `1` to `s.length - 1` (string is 1-indexed).
For example, strings **"abc ", "ac ", "b "** and **"abcbabcbcb "** are all happy strings and strings **"aa ", "baa "** and **"a... | null |
🔥 Pandas Step by Step and Concise Solution For Beginners 💯 | students-and-examinations | 0 | 1 | **\uD83D\uDD3C IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE \uD83D\uDC4D**\n```\nimport pandas as pd\n\ndef students_and_examinations(students: pd.DataFrame, subjects: pd.DataFrame, examinations: pd.DataFrame) -> pd.DataFrame:\n \n df1 = students.merge(subjects, how=\'cross\')\n \n df2 = examinations.groupby... | 7 | A **happy string** is a string that:
* consists only of letters of the set `['a', 'b', 'c']`.
* `s[i] != s[i + 1]` for all values of `i` from `1` to `s.length - 1` (string is 1-indexed).
For example, strings **"abc ", "ac ", "b "** and **"abcbabcbcb "** are all happy strings and strings **"aa ", "baa "** and **"a... | null |
Python simple code | subtract-the-product-and-sum-of-digits-of-an-integer | 0 | 1 | \n# Code\n```\nclass Solution:\n def subtractProductAndSum(self, n: int) -> int:\n s = 0\n mult = 1\n\n while n > 0:\n digit = n%10\n s += digit\n mult *= digit\n n = n//10\n\n return mult-s\n\n \n``` | 1 | Given an integer number `n`, return the difference between the product of its digits and the sum of its digits.
**Example 1:**
**Input:** n = 234
**Output:** 15
**Explanation:**
Product of digits = 2 \* 3 \* 4 = 24
Sum of digits = 2 + 3 + 4 = 9
Result = 24 - 9 = 15
**Example 2:**
**Input:** n = 4421
**Output:**... | Since we can rearrange the substring, all we care about is the frequency of each character in that substring. How to find the character frequencies efficiently ? As a preprocess, calculate the accumulate frequency of all characters for all prefixes of the string. How to check if a substring can be changed to a palindro... |
Python simple code | subtract-the-product-and-sum-of-digits-of-an-integer | 0 | 1 | \n# Code\n```\nclass Solution:\n def subtractProductAndSum(self, n: int) -> int:\n s = 0\n mult = 1\n\n while n > 0:\n digit = n%10\n s += digit\n mult *= digit\n n = n//10\n\n return mult-s\n\n \n``` | 1 | Alice and Bob continue their games with piles of stones. There are several stones **arranged in a row**, and each stone has an associated value which is an integer given in the array `stoneValue`.
Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take `1`, `2`, or `3` stones f... | How to compute all digits of the number ? Use modulus operator (%) to compute the last digit. Generalise modulus operator idea to compute all digits. |
Solution of subtract the product and sum of digits of an integer problem | subtract-the-product-and-sum-of-digits-of-an-integer | 0 | 1 | # Approach\n- Solved using simple loop\n- We iterate over a number and multiply and sum the digits of the number\n\n# Complexity\n- Time complexity:\n$$0(n)$$ - as we are iterating over n elements.\n\n# Code\n```\nclass Solution:\n def subtractProductAndSum(self, n: int) -> int:\n product_d, sum_d = 1, 0\n ... | 1 | Given an integer number `n`, return the difference between the product of its digits and the sum of its digits.
**Example 1:**
**Input:** n = 234
**Output:** 15
**Explanation:**
Product of digits = 2 \* 3 \* 4 = 24
Sum of digits = 2 + 3 + 4 = 9
Result = 24 - 9 = 15
**Example 2:**
**Input:** n = 4421
**Output:**... | Since we can rearrange the substring, all we care about is the frequency of each character in that substring. How to find the character frequencies efficiently ? As a preprocess, calculate the accumulate frequency of all characters for all prefixes of the string. How to check if a substring can be changed to a palindro... |
Solution of subtract the product and sum of digits of an integer problem | subtract-the-product-and-sum-of-digits-of-an-integer | 0 | 1 | # Approach\n- Solved using simple loop\n- We iterate over a number and multiply and sum the digits of the number\n\n# Complexity\n- Time complexity:\n$$0(n)$$ - as we are iterating over n elements.\n\n# Code\n```\nclass Solution:\n def subtractProductAndSum(self, n: int) -> int:\n product_d, sum_d = 1, 0\n ... | 1 | Alice and Bob continue their games with piles of stones. There are several stones **arranged in a row**, and each stone has an associated value which is an integer given in the array `stoneValue`.
Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take `1`, `2`, or `3` stones f... | How to compute all digits of the number ? Use modulus operator (%) to compute the last digit. Generalise modulus operator idea to compute all digits. |
Python | Easy Solution✅ | subtract-the-product-and-sum-of-digits-of-an-integer | 0 | 1 | ```\ndef subtractProductAndSum(self, n: int) -> int:\n prod = 1 # n =234\n sums = 0\n while n != 0: # 1st loop 2nd loop 3rd loop \n last = n % 10 # last = 4 3 2\n prod *= last # prod = 1*4 = 4 4*3 = 12 12*2 ... | 25 | Given an integer number `n`, return the difference between the product of its digits and the sum of its digits.
**Example 1:**
**Input:** n = 234
**Output:** 15
**Explanation:**
Product of digits = 2 \* 3 \* 4 = 24
Sum of digits = 2 + 3 + 4 = 9
Result = 24 - 9 = 15
**Example 2:**
**Input:** n = 4421
**Output:**... | Since we can rearrange the substring, all we care about is the frequency of each character in that substring. How to find the character frequencies efficiently ? As a preprocess, calculate the accumulate frequency of all characters for all prefixes of the string. How to check if a substring can be changed to a palindro... |
Python | Easy Solution✅ | subtract-the-product-and-sum-of-digits-of-an-integer | 0 | 1 | ```\ndef subtractProductAndSum(self, n: int) -> int:\n prod = 1 # n =234\n sums = 0\n while n != 0: # 1st loop 2nd loop 3rd loop \n last = n % 10 # last = 4 3 2\n prod *= last # prod = 1*4 = 4 4*3 = 12 12*2 ... | 25 | Alice and Bob continue their games with piles of stones. There are several stones **arranged in a row**, and each stone has an associated value which is an integer given in the array `stoneValue`.
Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take `1`, `2`, or `3` stones f... | How to compute all digits of the number ? Use modulus operator (%) to compute the last digit. Generalise modulus operator idea to compute all digits. |
Python one-liner using eval function (28ms, 14MB) | subtract-the-product-and-sum-of-digits-of-an-integer | 0 | 1 | Solution is relatively straightforward. You just have to make sure to cast `n` to be a string.\n\n```Python\ndef subtractProductAndSum(n)t:\n return eval(\'*\'.join(str(n))) - eval(\'+\'.join(str(n)))\n```\n\nThe time complexity of this solution is O(n) due to the two `join` operations.\n\n### Edit\nAs somebody said... | 68 | Given an integer number `n`, return the difference between the product of its digits and the sum of its digits.
**Example 1:**
**Input:** n = 234
**Output:** 15
**Explanation:**
Product of digits = 2 \* 3 \* 4 = 24
Sum of digits = 2 + 3 + 4 = 9
Result = 24 - 9 = 15
**Example 2:**
**Input:** n = 4421
**Output:**... | Since we can rearrange the substring, all we care about is the frequency of each character in that substring. How to find the character frequencies efficiently ? As a preprocess, calculate the accumulate frequency of all characters for all prefixes of the string. How to check if a substring can be changed to a palindro... |
Python one-liner using eval function (28ms, 14MB) | subtract-the-product-and-sum-of-digits-of-an-integer | 0 | 1 | Solution is relatively straightforward. You just have to make sure to cast `n` to be a string.\n\n```Python\ndef subtractProductAndSum(n)t:\n return eval(\'*\'.join(str(n))) - eval(\'+\'.join(str(n)))\n```\n\nThe time complexity of this solution is O(n) due to the two `join` operations.\n\n### Edit\nAs somebody said... | 68 | Alice and Bob continue their games with piles of stones. There are several stones **arranged in a row**, and each stone has an associated value which is an integer given in the array `stoneValue`.
Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take `1`, `2`, or `3` stones f... | How to compute all digits of the number ? Use modulus operator (%) to compute the last digit. Generalise modulus operator idea to compute all digits. |
Python ( dict ) and C++ ( map ) | hashtable with greedy | group-the-people-given-the-group-size-they-belong-to | 0 | 1 | ```CPP []\nclass Solution {\npublic:\n vector<vector<int>> groupThePeople(vector<int>& gS) {\n map<int,vector<int>>m;\n int i = 0;\n for(auto &sz: gS) m[sz].push_back(i++);\n vector<vector<int>>ans;\n for(auto &i: m){\n vector<int>temp;\n for(int j = 0; j < i.... | 2 | There are `n` people that are split into some unknown number of groups. Each person is labeled with a **unique ID** from `0` to `n - 1`.
You are given an integer array `groupSizes`, where `groupSizes[i]` is the size of the group that person `i` is in. For example, if `groupSizes[1] = 3`, then person `1` must be in a g... | Exploit the fact that the length of the puzzle is only 7. Use bit-masks to represent the word and puzzle strings. For each puzzle, count the number of words whose bit-mask is a sub-mask of the puzzle's bit-mask. |
Python3 👍||⚡94% faster beats, only 10 lines🔥|| clean solution || | group-the-people-given-the-group-size-they-belong-to | 0 | 1 | \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. $$O(n)$$ -->\n\n# Code\n```\nclass Solut... | 2 | There are `n` people that are split into some unknown number of groups. Each person is labeled with a **unique ID** from `0` to `n - 1`.
You are given an integer array `groupSizes`, where `groupSizes[i]` is the size of the group that person `i` is in. For example, if `groupSizes[1] = 3`, then person `1` must be in a g... | Exploit the fact that the length of the puzzle is only 7. Use bit-masks to represent the word and puzzle strings. For each puzzle, count the number of words whose bit-mask is a sub-mask of the puzzle's bit-mask. |
simple python3 solution (list comprehension, python list slicing) | group-the-people-given-the-group-size-they-belong-to | 0 | 1 | # Intuition\nJust track the person index and his belong **groupsize** then later he has to be grouped into a group with other person with same **groupsize** while maintaining each group size by using python list *slicing* method.\n\n# Complexity\n- Time complexity:\nO(n): one loop is to go through the __groupSizes__ l... | 2 | There are `n` people that are split into some unknown number of groups. Each person is labeled with a **unique ID** from `0` to `n - 1`.
You are given an integer array `groupSizes`, where `groupSizes[i]` is the size of the group that person `i` is in. For example, if `groupSizes[1] = 3`, then person `1` must be in a g... | Exploit the fact that the length of the puzzle is only 7. Use bit-masks to represent the word and puzzle strings. For each puzzle, count the number of words whose bit-mask is a sub-mask of the puzzle's bit-mask. |
Python short and clean. Functional programming. | group-the-people-given-the-group-size-they-belong-to | 0 | 1 | # Approach\n1. Create a HashMap called `groups` that maps from `group_size` to list of `people_ids` belonging to that group.\n\n2. For each `group_size` in `groups`, divide the `people_ids` sequence into sub sequences of length `group_size`.\nThis can be done either while inserting to `groups` or as a seperate operatio... | 1 | There are `n` people that are split into some unknown number of groups. Each person is labeled with a **unique ID** from `0` to `n - 1`.
You are given an integer array `groupSizes`, where `groupSizes[i]` is the size of the group that person `i` is in. For example, if `groupSizes[1] = 3`, then person `1` must be in a g... | Exploit the fact that the length of the puzzle is only 7. Use bit-masks to represent the word and puzzle strings. For each puzzle, count the number of words whose bit-mask is a sub-mask of the puzzle's bit-mask. |
Python3, C++ | Brute-force -> Optimal | Full Explanation | find-the-smallest-divisor-given-a-threshold | 0 | 1 | ### Find the Smallest Divisor Given a Threshold\n\nGiven an array of integers `nums` and an integer `threshold`, we will choose a positive integer `divisor`, divide all the array by it, and sum the division\'s result. Find the **smallest** `divisor` such that the result mentioned above is less than or equal to `thresho... | 1 | Given an array of integers `nums` and an integer `threshold`, we will choose a positive integer `divisor`, divide all the array by it, and sum the division's result. Find the **smallest** `divisor` such that the result mentioned above is less than or equal to `threshold`.
Each result of the division is rounded to the ... | Handle the conversions of day, month and year separately. Notice that days always have a two-word ending, so if you erase the last two characters of this days you'll get the number. |
Python3, C++ | Brute-force -> Optimal | Full Explanation | find-the-smallest-divisor-given-a-threshold | 0 | 1 | ### Find the Smallest Divisor Given a Threshold\n\nGiven an array of integers `nums` and an integer `threshold`, we will choose a positive integer `divisor`, divide all the array by it, and sum the division\'s result. Find the **smallest** `divisor` such that the result mentioned above is less than or equal to `thresho... | 1 | Given an array of string `words`, return _all strings in_ `words` _that is a **substring** of another word_. You can return the answer in **any order**.
A **substring** is a contiguous sequence of characters within a string
**Example 1:**
**Input:** words = \[ "mass ", "as ", "hero ", "superhero "\]
**Output:** \[ "... | Examine every possible number for solution. Choose the largest of them. Use binary search to reduce the time complexity. |
Simple Binary Search Approach in Python | find-the-smallest-divisor-given-a-threshold | 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 Binary Search \n\n# Complexity\n- Time complexity: O( log( max(array) ) )\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1) constant ... | 4 | Given an array of integers `nums` and an integer `threshold`, we will choose a positive integer `divisor`, divide all the array by it, and sum the division's result. Find the **smallest** `divisor` such that the result mentioned above is less than or equal to `threshold`.
Each result of the division is rounded to the ... | Handle the conversions of day, month and year separately. Notice that days always have a two-word ending, so if you erase the last two characters of this days you'll get the number. |
Simple Binary Search Approach in Python | find-the-smallest-divisor-given-a-threshold | 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 Binary Search \n\n# Complexity\n- Time complexity: O( log( max(array) ) )\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1) constant ... | 4 | Given an array of string `words`, return _all strings in_ `words` _that is a **substring** of another word_. You can return the answer in **any order**.
A **substring** is a contiguous sequence of characters within a string
**Example 1:**
**Input:** words = \[ "mass ", "as ", "hero ", "superhero "\]
**Output:** \[ "... | Examine every possible number for solution. Choose the largest of them. Use binary search to reduce the time complexity. |
Python BFS & Bit operation | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | \nclass Solution:\n\n def minFlips(self, mat: List[List[int]]) -> int:\n m,n=len(mat),len(mat[0])\n def convert_to_bit(M):\n start=0\n for i in range(m):\n for j in range(n):\n start|=M[i][j]<<(i*n+j)\n return start\n init=conver... | 3 | Given a `m x n` binary matrix `mat`. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing `1` to `0` and `0` to `1`). A pair of cells are called neighbors if they share one edge.
Return the _minimum number of steps_ required to convert `mat` to a zero matrix... | Find the divisors of each element in the array. You only need to loop to the square root of a number to find its divisors. |
Python BFS & Bit operation | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | \nclass Solution:\n\n def minFlips(self, mat: List[List[int]]) -> int:\n m,n=len(mat),len(mat[0])\n def convert_to_bit(M):\n start=0\n for i in range(m):\n for j in range(n):\n start|=M[i][j]<<(i*n+j)\n return start\n init=conver... | 3 | Given the array `queries` of positive integers between `1` and `m`, you have to process all `queries[i]` (from `i=0` to `i=queries.length-1`) according to the following rules:
* In the beginning, you have the permutation `P=[1,2,3,...,m]`.
* For the current `i`, find the position of `queries[i]` in the permutation... | Flipping same index two times is like not flipping it at all. Each index can be flipped one time. Try all possible combinations. O(2^(n*m)). |
BFS + Bit manipulation Python3 Solution | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | ```\nclass Solution:\n \n def getFlattnedIndex(self, row: int, col: int, cols: int) -> int:\n return row*cols + col\n \n \n def getNeighbors(self, row: int, col: int, mat: List[List[int]]) -> List[Tuple[int]]:\n \n nbrs = []\n directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]\n ... | 0 | Given a `m x n` binary matrix `mat`. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing `1` to `0` and `0` to `1`). A pair of cells are called neighbors if they share one edge.
Return the _minimum number of steps_ required to convert `mat` to a zero matrix... | Find the divisors of each element in the array. You only need to loop to the square root of a number to find its divisors. |
BFS + Bit manipulation Python3 Solution | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | ```\nclass Solution:\n \n def getFlattnedIndex(self, row: int, col: int, cols: int) -> int:\n return row*cols + col\n \n \n def getNeighbors(self, row: int, col: int, mat: List[List[int]]) -> List[Tuple[int]]:\n \n nbrs = []\n directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]\n ... | 0 | Given the array `queries` of positive integers between `1` and `m`, you have to process all `queries[i]` (from `i=0` to `i=queries.length-1`) according to the following rules:
* In the beginning, you have the permutation `P=[1,2,3,...,m]`.
* For the current `i`, find the position of `queries[i]` in the permutation... | Flipping same index two times is like not flipping it at all. Each index can be flipped one time. Try all possible combinations. O(2^(n*m)). |
Python Hard | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | ```\nclass Solution:\n def minFlips(self, mat: List[List[int]]) -> int:\n \'\'\'\n try bfs + heap\n \'\'\'\n directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]\n M, N = len(mat), len(mat[0])\n h = []\n\n seen = set()\n\n heapq.heappush(h, (0, mat))\n\n whil... | 0 | Given a `m x n` binary matrix `mat`. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing `1` to `0` and `0` to `1`). A pair of cells are called neighbors if they share one edge.
Return the _minimum number of steps_ required to convert `mat` to a zero matrix... | Find the divisors of each element in the array. You only need to loop to the square root of a number to find its divisors. |
Python Hard | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | ```\nclass Solution:\n def minFlips(self, mat: List[List[int]]) -> int:\n \'\'\'\n try bfs + heap\n \'\'\'\n directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]\n M, N = len(mat), len(mat[0])\n h = []\n\n seen = set()\n\n heapq.heappush(h, (0, mat))\n\n whil... | 0 | Given the array `queries` of positive integers between `1` and `m`, you have to process all `queries[i]` (from `i=0` to `i=queries.length-1`) according to the following rules:
* In the beginning, you have the permutation `P=[1,2,3,...,m]`.
* For the current `i`, find the position of `queries[i]` in the permutation... | Flipping same index two times is like not flipping it at all. Each index can be flipped one time. Try all possible combinations. O(2^(n*m)). |
Python3 - Bitwise operations | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | # Code\n```\nclass Solution:\n def minFlips(self, mat: List[List[int]]) -> int:\n mx, nx = len(mat), len(mat[0])\n bv = 1\n m = defaultdict(int)\n \n c = 0\n for i in range(mx):\n for j in range(nx):\n m[(i, j)] = bv\n \n ... | 0 | Given a `m x n` binary matrix `mat`. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing `1` to `0` and `0` to `1`). A pair of cells are called neighbors if they share one edge.
Return the _minimum number of steps_ required to convert `mat` to a zero matrix... | Find the divisors of each element in the array. You only need to loop to the square root of a number to find its divisors. |
Python3 - Bitwise operations | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | # Code\n```\nclass Solution:\n def minFlips(self, mat: List[List[int]]) -> int:\n mx, nx = len(mat), len(mat[0])\n bv = 1\n m = defaultdict(int)\n \n c = 0\n for i in range(mx):\n for j in range(nx):\n m[(i, j)] = bv\n \n ... | 0 | Given the array `queries` of positive integers between `1` and `m`, you have to process all `queries[i]` (from `i=0` to `i=queries.length-1`) according to the following rules:
* In the beginning, you have the permutation `P=[1,2,3,...,m]`.
* For the current `i`, find the position of `queries[i]` in the permutation... | Flipping same index two times is like not flipping it at all. Each index can be flipped one time. Try all possible combinations. O(2^(n*m)). |
Concise solution on Python3 | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | # Code\n```\nclass Solution:\n def minFlips(self, mat: List[List[int]]) -> int:\n\n n = len(mat)\n m = len(mat[0])\n\n def ins(i, j):\n return 0 <= i < n and 0 <= j < m\n\n def swipe(i, j):\n mat[i][j] ^= 1\n\n if ins(i-1, j):\n mat[i-1][j] ... | 0 | Given a `m x n` binary matrix `mat`. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing `1` to `0` and `0` to `1`). A pair of cells are called neighbors if they share one edge.
Return the _minimum number of steps_ required to convert `mat` to a zero matrix... | Find the divisors of each element in the array. You only need to loop to the square root of a number to find its divisors. |
Concise solution on Python3 | minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix | 0 | 1 | # Code\n```\nclass Solution:\n def minFlips(self, mat: List[List[int]]) -> int:\n\n n = len(mat)\n m = len(mat[0])\n\n def ins(i, j):\n return 0 <= i < n and 0 <= j < m\n\n def swipe(i, j):\n mat[i][j] ^= 1\n\n if ins(i-1, j):\n mat[i-1][j] ... | 0 | Given the array `queries` of positive integers between `1` and `m`, you have to process all `queries[i]` (from `i=0` to `i=queries.length-1`) according to the following rules:
* In the beginning, you have the permutation `P=[1,2,3,...,m]`.
* For the current `i`, find the position of `queries[i]` in the permutation... | Flipping same index two times is like not flipping it at all. Each index can be flipped one time. Try all possible combinations. O(2^(n*m)). |
Python super-simple and easy solution | iterator-for-combination | 0 | 1 | ```\nclass CombinationIterator:\n def __init__(self, characters: str, combinationLength: int):\n self.allCombinations = list(combinations(characters, combinationLength))\n self.count = 0\n\n def next(self) -> str:\n self.count += 1\n return "".join(self.allCombinations[self.count-1])\n... | 7 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
Simple python solution for iterator if combination | iterator-for-combination | 0 | 1 | 1) Create a list of combinations using binary numbers.\n\te.g. In "abc", there are two possibilities for each character x : include it or not \n\t\t\tTherefore, we may have [1,0,1] or [0,1,1] or [1,1,0] or [0,0,0] or [1,1,1],etc \n\t\t\t\n2) Check if number of ones in each possible binary list is equal to combination l... | 6 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
Python || Backtracking | iterator-for-combination | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse backtracking (comb method from the code) to find all the combinations of length combinationLength from characters. As you find next, pop from the list.\n\n\n# Code\n```\nclass CombinationIterator:\n def __init__(self, characters: s... | 0 | Design the `CombinationIterator` class:
* `CombinationIterator(string characters, int combinationLength)` Initializes the object with a string `characters` of **sorted distinct** lowercase English letters and a number `combinationLength` as arguments.
* `next()` Returns the next combination of length `combinationL... | Use dynamic programming. Let dp[i] be the solution for the prefix of the array that ends at index i, if the element at index i is in the subsequence. dp[i] = nums[i] + max(0, dp[i-k], dp[i-k+1], ..., dp[i-1]) Use a heap with the sliding window technique to optimize the dp. |
C++/Python frequency count vs Binary search||0ms Beats 100% | element-appearing-more-than-25-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n3 different kinds of soultion arte provided.\n1 is very naive, solved this with counting frequency for each number; no hash table is used.\nThe 2nd approach uses binary search if the possible target \n`int x[5]={arr[0], arr[n/4], arr[n/2]... | 11 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
C++/Python frequency count vs Binary search||0ms Beats 100% | element-appearing-more-than-25-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n3 different kinds of soultion arte provided.\n1 is very naive, solved this with counting frequency for each number; no hash table is used.\nThe 2nd approach uses binary search if the possible target \n`int x[5]={arr[0], arr[n/4], arr[n/2]... | 11 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
【Video】Give me 5 minutes - 2 solutions - How we think about a solution | element-appearing-more-than-25-in-sorted-array | 1 | 1 | # Intuition\nCalculate quarter length of input array.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/bTm-6y7Ob0A\n\n\u25A0 Timeline of the video\n\n`0:07` Explain algorithm of Solution 1\n`2:46` Coding of solution 1\n`3:49` Time Complexity and Space Complexity of solution 1\n`4:02` Step by step algorithm of solution 1\... | 78 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
【Video】Give me 5 minutes - 2 solutions - How we think about a solution | element-appearing-more-than-25-in-sorted-array | 1 | 1 | # Intuition\nCalculate quarter length of input array.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/bTm-6y7Ob0A\n\n\u25A0 Timeline of the video\n\n`0:07` Explain algorithm of Solution 1\n`2:46` Coding of solution 1\n`3:49` Time Complexity and Space Complexity of solution 1\n`4:02` Step by step algorithm of solution 1\... | 78 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
Simple approach without use of hashmap and extra libraries! | element-appearing-more-than-25-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSince the array elements are sorted whenever in the loop ith element equals i+len(arr)//4 th element. We will return the element.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!... | 3 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
Simple approach without use of hashmap and extra libraries! | element-appearing-more-than-25-in-sorted-array | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSince the array elements are sorted whenever in the loop ith element equals i+len(arr)//4 th element. We will return the element.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!... | 3 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
⭐One-Liner Hack in 🐍|| Two Approaches ||😐Beats 75.32%😐⭐ | element-appearing-more-than-25-in-sorted-array | 0 | 1 | \n# Code\n```py\nclass Solution:\n def findSpecialInteger(self, arr: List[int]) -> int:\n return next(i for i in arr if arr.count(i) / len(arr) > 0.25)\n```\n ---\n\n\n# Code\n```py\nclass Solution... | 3 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
⭐One-Liner Hack in 🐍|| Two Approaches ||😐Beats 75.32%😐⭐ | element-appearing-more-than-25-in-sorted-array | 0 | 1 | \n# Code\n```py\nclass Solution:\n def findSpecialInteger(self, arr: List[int]) -> int:\n return next(i for i in arr if arr.count(i) / len(arr) > 0.25)\n```\n ---\n\n\n# Code\n```py\nclass Solution... | 3 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
Beats 97% very Easy | element-appearing-more-than-25-in-sorted-array | 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. $$O(n)$$ --... | 8 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
Beats 97% very Easy | element-appearing-more-than-25-in-sorted-array | 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. $$O(n)$$ --... | 8 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
Best and Easy Approach...... || Beginners friendly...🔥🔥🔥 | element-appearing-more-than-25-in-sorted-array | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialize an empty dictionary \'arr2\' to store the count of occurrences for each unique element in the input array.\n2. Iterate through each element i in the input array arr.\n3. If i is already a key in \'arr2\', increment its count.\n4. If i is... | 2 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
Best and Easy Approach...... || Beginners friendly...🔥🔥🔥 | element-appearing-more-than-25-in-sorted-array | 0 | 1 | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialize an empty dictionary \'arr2\' to store the count of occurrences for each unique element in the input array.\n2. Iterate through each element i in the input array arr.\n3. If i is already a key in \'arr2\', increment its count.\n4. If i is... | 2 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
✅✅Majority Element in Array🔥🔥 | element-appearing-more-than-25-in-sorted-array | 1 | 1 | # Intuition\nThe problem seems to ask for finding a "special" integer, which appears more than 25% of the time in the given array. The approach appears to use a dictionary to count the occurrences of each element and then iterate through the dictionary to find the element that meets the criteria.\n\n# Approach\n1. Chec... | 11 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
✅✅Majority Element in Array🔥🔥 | element-appearing-more-than-25-in-sorted-array | 1 | 1 | # Intuition\nThe problem seems to ask for finding a "special" integer, which appears more than 25% of the time in the given array. The approach appears to use a dictionary to count the occurrences of each element and then iterate through the dictionary to find the element that meets the criteria.\n\n# Approach\n1. Chec... | 11 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
Easiest Solution || Python | element-appearing-more-than-25-in-sorted-array | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def findSpecialInteger(self, arr: List[int]) -> int:\n n= len(arr)\n threshold= n//4\n\n candidate=arr[0]\n count = 1\n for i in range(1,n):\n if arr[i] == candidate:\n count+=1\n else:\n count =... | 1 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
Easiest Solution || Python | element-appearing-more-than-25-in-sorted-array | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def findSpecialInteger(self, arr: List[int]) -> int:\n n= len(arr)\n threshold= n//4\n\n candidate=arr[0]\n count = 1\n for i in range(1,n):\n if arr[i] == candidate:\n count+=1\n else:\n count =... | 1 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
Using Counter in python | element-appearing-more-than-25-in-sorted-array | 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. $$O(n)$$ -->Linear O(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. ... | 1 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
Using Counter in python | element-appearing-more-than-25-in-sorted-array | 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. $$O(n)$$ -->Linear O(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. ... | 1 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
Solution of element appearing more than 25 in sorted array problem | element-appearing-more-than-25-in-sorted-array | 0 | 1 | # Approach\n1. Step one: Compare the current number (`arr[idx]`) with the previous number (`cur_number`)\n- If they are equal, increment the count `count`.\n- If they are not equal, reset the count `count` to 1.\n2. Step two: Check if the count `count` is greater than 25% of the \narray size (`0.25 * len(arr)`)\n3. Ste... | 1 | Given an integer array **sorted** in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.
**Example 1:**
**Input:** arr = \[1,2,2,6,6,6,6,7,10\]
**Output:** 6
**Example 2:**
**Input:** arr = \[1,1\]
**Output:** 1
**Constraints:**
* `1 <= arr... | Find the distance between the two stops if the bus moved in clockwise or counterclockwise directions. |
Solution of element appearing more than 25 in sorted array problem | element-appearing-more-than-25-in-sorted-array | 0 | 1 | # Approach\n1. Step one: Compare the current number (`arr[idx]`) with the previous number (`cur_number`)\n- If they are equal, increment the count `count`.\n- If they are not equal, reset the count `count` to 1.\n2. Step two: Check if the count `count` is greater than 25% of the \narray size (`0.25 * len(arr)`)\n3. Ste... | 1 | **Balanced** strings are those that have an equal quantity of `'L'` and `'R'` characters.
Given a **balanced** string `s`, split it into some number of substrings such that:
* Each substring is balanced.
Return _the **maximum** number of balanced strings you can obtain._
**Example 1:**
**Input:** s = "RLRRLLRLR... | Divide the array in four parts [1 - 25%] [25 - 50 %] [50 - 75 %] [75% - 100%] The answer should be in one of the ends of the intervals. In order to check which is element is the answer we can count the frequency with binarySearch. |
✔️ [Python3] SORTING 👀, Explained | remove-covered-intervals | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nAs for almost all problems related to intervals, we have to sort them first by the starting position. The next problem is how to define whether an interval is covered? After the sorting, we know that the start of eve... | 49 | Given an array `intervals` where `intervals[i] = [li, ri]` represent the interval `[li, ri)`, remove all intervals that are covered by another interval in the list.
The interval `[a, b)` is covered by the interval `[c, d)` if and only if `c <= a` and `b <= d`.
Return _the number of remaining intervals_.
**Example 1:... | How to solve this problem if no deletions are allowed ? Try deleting each element and find the maximum subarray sum to both sides of that element. To do that efficiently, use the idea of Kadane's algorithm. |
✔️ [Python3] SORTING 👀, Explained | remove-covered-intervals | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nAs for almost all problems related to intervals, we have to sort them first by the starting position. The next problem is how to define whether an interval is covered? After the sorting, we know that the start of eve... | 49 | On a **0-indexed** `8 x 8` chessboard, there can be multiple black queens ad one white king.
You are given a 2D integer array `queens` where `queens[i] = [xQueeni, yQueeni]` represents the position of the `ith` black queen on the chessboard. You are also given an integer array `king` of length `2` where `king = [xKing... | How to check if an interval is covered by another? Compare each interval to all others and check if it is covered by any interval. |
Python Simple Solution Explained (video + code) (Fastest) | remove-covered-intervals | 0 | 1 | [](https://www.youtube.com/watch?v=emPnw5m2nN0)\nhttps://www.youtube.com/watch?v=emPnw5m2nN0\n```\nclass Solution:\n def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:\n intervals = sorted(intervals, key = lambda x : (x[0], -x[1]))\n \n res = 0\n ending = 0\n \n ... | 14 | Given an array `intervals` where `intervals[i] = [li, ri]` represent the interval `[li, ri)`, remove all intervals that are covered by another interval in the list.
The interval `[a, b)` is covered by the interval `[c, d)` if and only if `c <= a` and `b <= d`.
Return _the number of remaining intervals_.
**Example 1:... | How to solve this problem if no deletions are allowed ? Try deleting each element and find the maximum subarray sum to both sides of that element. To do that efficiently, use the idea of Kadane's algorithm. |
Python Simple Solution Explained (video + code) (Fastest) | remove-covered-intervals | 0 | 1 | [](https://www.youtube.com/watch?v=emPnw5m2nN0)\nhttps://www.youtube.com/watch?v=emPnw5m2nN0\n```\nclass Solution:\n def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:\n intervals = sorted(intervals, key = lambda x : (x[0], -x[1]))\n \n res = 0\n ending = 0\n \n ... | 14 | On a **0-indexed** `8 x 8` chessboard, there can be multiple black queens ad one white king.
You are given a 2D integer array `queens` where `queens[i] = [xQueeni, yQueeni]` represents the position of the `ith` black queen on the chessboard. You are also given an integer array `king` of length `2` where `king = [xKing... | How to check if an interval is covered by another? Compare each interval to all others and check if it is covered by any interval. |
[ Python ] ✔✔ Simple Python Solution By Sorting the List 🔥✌ | remove-covered-intervals | 0 | 1 | # If It is Useful to Understand Please Upvote Me \uD83D\uDE4F\uD83D\uDE4F\n\tclass Solution:\n\t\tdef removeCoveredIntervals(self, intervals: List[List[int]]) -> int:\n\n\t\t\tintervals=sorted(intervals)\n\n\t\t\ti=0\n\t\t\twhile i<len(intervals)-1:\n\n\t\t\t\t\ta,b = intervals[i]\n\t\t\t\t\tp,q = intervals[i+1]\n\n\t\... | 10 | Given an array `intervals` where `intervals[i] = [li, ri]` represent the interval `[li, ri)`, remove all intervals that are covered by another interval in the list.
The interval `[a, b)` is covered by the interval `[c, d)` if and only if `c <= a` and `b <= d`.
Return _the number of remaining intervals_.
**Example 1:... | How to solve this problem if no deletions are allowed ? Try deleting each element and find the maximum subarray sum to both sides of that element. To do that efficiently, use the idea of Kadane's algorithm. |
[ Python ] ✔✔ Simple Python Solution By Sorting the List 🔥✌ | remove-covered-intervals | 0 | 1 | # If It is Useful to Understand Please Upvote Me \uD83D\uDE4F\uD83D\uDE4F\n\tclass Solution:\n\t\tdef removeCoveredIntervals(self, intervals: List[List[int]]) -> int:\n\n\t\t\tintervals=sorted(intervals)\n\n\t\t\ti=0\n\t\t\twhile i<len(intervals)-1:\n\n\t\t\t\t\ta,b = intervals[i]\n\t\t\t\t\tp,q = intervals[i+1]\n\n\t\... | 10 | On a **0-indexed** `8 x 8` chessboard, there can be multiple black queens ad one white king.
You are given a 2D integer array `queens` where `queens[i] = [xQueeni, yQueeni]` represents the position of the `ith` black queen on the chessboard. You are also given an integer array `king` of length `2` where `king = [xKing... | How to check if an interval is covered by another? Compare each interval to all others and check if it is covered by any interval. |
Python3 || brute force - 40% Faster | remove-covered-intervals | 0 | 1 | ```\nclass Solution:\n def removeCoveredIntervals(self, new: List[List[int]]) -> int:\n arr=[]\n for i in range(len(new)):\n for j in range(len(new)):\n if i!=j and new[j][0] <= new[i][0] and new[i][1] <= new[j][1]:\n arr.append(new[i])\n ... | 3 | Given an array `intervals` where `intervals[i] = [li, ri]` represent the interval `[li, ri)`, remove all intervals that are covered by another interval in the list.
The interval `[a, b)` is covered by the interval `[c, d)` if and only if `c <= a` and `b <= d`.
Return _the number of remaining intervals_.
**Example 1:... | How to solve this problem if no deletions are allowed ? Try deleting each element and find the maximum subarray sum to both sides of that element. To do that efficiently, use the idea of Kadane's algorithm. |
Python3 || brute force - 40% Faster | remove-covered-intervals | 0 | 1 | ```\nclass Solution:\n def removeCoveredIntervals(self, new: List[List[int]]) -> int:\n arr=[]\n for i in range(len(new)):\n for j in range(len(new)):\n if i!=j and new[j][0] <= new[i][0] and new[i][1] <= new[j][1]:\n arr.append(new[i])\n ... | 3 | On a **0-indexed** `8 x 8` chessboard, there can be multiple black queens ad one white king.
You are given a 2D integer array `queens` where `queens[i] = [xQueeni, yQueeni]` represents the position of the `ith` black queen on the chessboard. You are also given an integer array `king` of length `2` where `king = [xKing... | How to check if an interval is covered by another? Compare each interval to all others and check if it is covered by any interval. |
Python 3 (90ms) | Sorted Matrix Solution | Easy to Understand | remove-covered-intervals | 0 | 1 | ```\nclass Solution:\n def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:\n intervals = sorted(intervals, key = lambda x : (x[0], -x[1]))\n res = 0\n ending = 0\n for _, end in intervals:\n if end > ending:\n res += 1\n ending = e... | 3 | Given an array `intervals` where `intervals[i] = [li, ri]` represent the interval `[li, ri)`, remove all intervals that are covered by another interval in the list.
The interval `[a, b)` is covered by the interval `[c, d)` if and only if `c <= a` and `b <= d`.
Return _the number of remaining intervals_.
**Example 1:... | How to solve this problem if no deletions are allowed ? Try deleting each element and find the maximum subarray sum to both sides of that element. To do that efficiently, use the idea of Kadane's algorithm. |
Python 3 (90ms) | Sorted Matrix Solution | Easy to Understand | remove-covered-intervals | 0 | 1 | ```\nclass Solution:\n def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:\n intervals = sorted(intervals, key = lambda x : (x[0], -x[1]))\n res = 0\n ending = 0\n for _, end in intervals:\n if end > ending:\n res += 1\n ending = e... | 3 | On a **0-indexed** `8 x 8` chessboard, there can be multiple black queens ad one white king.
You are given a 2D integer array `queens` where `queens[i] = [xQueeni, yQueeni]` represents the position of the `ith` black queen on the chessboard. You are also given an integer array `king` of length `2` where `king = [xKing... | How to check if an interval is covered by another? Compare each interval to all others and check if it is covered by any interval. |
Python | Faster than 99% | Using Dict. | remove-covered-intervals | 0 | 1 | The idea is to keep a track of the highest ending time for every starting time and storing it as a key.\nEx - [[1,2],[1,4],[2,6]]\nwe would store : {1:4 , 2:6}\nWe only need the max ones becuase the ones lesser than that will be included in the highest one anyway.\n\nLike in our example, [1,2] is in [1,4] that\'s why w... | 6 | Given an array `intervals` where `intervals[i] = [li, ri]` represent the interval `[li, ri)`, remove all intervals that are covered by another interval in the list.
The interval `[a, b)` is covered by the interval `[c, d)` if and only if `c <= a` and `b <= d`.
Return _the number of remaining intervals_.
**Example 1:... | How to solve this problem if no deletions are allowed ? Try deleting each element and find the maximum subarray sum to both sides of that element. To do that efficiently, use the idea of Kadane's algorithm. |
Python | Faster than 99% | Using Dict. | remove-covered-intervals | 0 | 1 | The idea is to keep a track of the highest ending time for every starting time and storing it as a key.\nEx - [[1,2],[1,4],[2,6]]\nwe would store : {1:4 , 2:6}\nWe only need the max ones becuase the ones lesser than that will be included in the highest one anyway.\n\nLike in our example, [1,2] is in [1,4] that\'s why w... | 6 | On a **0-indexed** `8 x 8` chessboard, there can be multiple black queens ad one white king.
You are given a 2D integer array `queens` where `queens[i] = [xQueeni, yQueeni]` represents the position of the `ith` black queen on the chessboard. You are also given an integer array `king` of length `2` where `king = [xKing... | How to check if an interval is covered by another? Compare each interval to all others and check if it is covered by any interval. |
python easy DP beats 98.27% | minimum-falling-path-sum-ii | 0 | 1 | \n\n```\ndef minFallingPathSum(self, g: List[List[int]]) -> int:\n l=len(g)\n for i in range(1,l):\n temp = sorted(g[i-1])\n for j in range(l):\n if g[i-1][j]... | 1 | Given an `n x n` integer matrix `grid`, return _the minimum sum of a **falling path with non-zero shifts**_.
A **falling path with non-zero shifts** is a choice of exactly one element from each row of `grid` such that no two elements chosen in adjacent rows are in the same column.
**Example 1:**
**Input:** arr = \[\... | Sum up the number of days for the years before the given year. Handle the case of a leap year. Find the number of days for each month of the given year. |
python easy DP beats 98.27% | minimum-falling-path-sum-ii | 0 | 1 | \n\n```\ndef minFallingPathSum(self, g: List[List[int]]) -> int:\n l=len(g)\n for i in range(1,l):\n temp = sorted(g[i-1])\n for j in range(l):\n if g[i-1][j]... | 1 | Given an array `nums` of positive integers, return the longest possible length of an array prefix of `nums`, such that it is possible to remove **exactly one** element from this prefix so that every number that has appeared in it will have the same number of occurrences.
If after removing one element there are no rema... | Use dynamic programming. Let dp[i][j] be the answer for the first i rows such that column j is chosen from row i. Use the concept of cumulative array to optimize the complexity of the solution. |
🐍 Python3 Clean Solution using DP | minimum-falling-path-sum-ii | 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. $$O(n)$$ --... | 1 | Given an `n x n` integer matrix `grid`, return _the minimum sum of a **falling path with non-zero shifts**_.
A **falling path with non-zero shifts** is a choice of exactly one element from each row of `grid` such that no two elements chosen in adjacent rows are in the same column.
**Example 1:**
**Input:** arr = \[\... | Sum up the number of days for the years before the given year. Handle the case of a leap year. Find the number of days for each month of the given year. |
🐍 Python3 Clean Solution using DP | minimum-falling-path-sum-ii | 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. $$O(n)$$ --... | 1 | Given an array `nums` of positive integers, return the longest possible length of an array prefix of `nums`, such that it is possible to remove **exactly one** element from this prefix so that every number that has appeared in it will have the same number of occurrences.
If after removing one element there are no rema... | Use dynamic programming. Let dp[i][j] be the answer for the first i rows such that column j is chosen from row i. Use the concept of cumulative array to optimize the complexity of the solution. |
hi | convert-binary-number-in-a-linked-list-to-integer | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nhi\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nlinear\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\no(n)\n\n- Space complexity:\n<!-- Add your space complexity here... | 0 | Given `head` which is a reference node to a singly-linked list. The value of each node in the linked list is either `0` or `1`. The linked list holds the binary representation of a number.
Return the _decimal value_ of the number in the linked list.
The **most significant bit** is at the head of the linked list.
**E... | Use dynamic programming. The state would be the index in arr1 and the index of the previous element in arr2 after sorting it and removing duplicates. |
hi | convert-binary-number-in-a-linked-list-to-integer | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nhi\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nlinear\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\no(n)\n\n- Space complexity:\n<!-- Add your space complexity here... | 0 | You have a `grid` of size `n x 3` and you want to paint each cell of the grid with exactly one of the three colors: **Red**, **Yellow,** or **Green** while making sure that no two adjacent cells have the same color (i.e., no two cells that share vertical or horizontal sides have the same color).
Given `n` the number o... | Traverse the linked list and store all values in a string or array. convert the values obtained to decimal value. You can solve the problem in O(1) memory using bits operation. use shift left operation ( << ) and or operation ( | ) to get the decimal value in one operation. |
Nurbek | convert-binary-number-in-a-linked-list-to-integer | 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 `head` which is a reference node to a singly-linked list. The value of each node in the linked list is either `0` or `1`. The linked list holds the binary representation of a number.
Return the _decimal value_ of the number in the linked list.
The **most significant bit** is at the head of the linked list.
**E... | Use dynamic programming. The state would be the index in arr1 and the index of the previous element in arr2 after sorting it and removing duplicates. |
Nurbek | convert-binary-number-in-a-linked-list-to-integer | 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 | You have a `grid` of size `n x 3` and you want to paint each cell of the grid with exactly one of the three colors: **Red**, **Yellow,** or **Green** while making sure that no two adjacent cells have the same color (i.e., no two cells that share vertical or horizontal sides have the same color).
Given `n` the number o... | Traverse the linked list and store all values in a string or array. convert the values obtained to decimal value. You can solve the problem in O(1) memory using bits operation. use shift left operation ( << ) and or operation ( | ) to get the decimal value in one operation. |
[Python] Simple. 20ms. | convert-binary-number-in-a-linked-list-to-integer | 0 | 1 | Read the binary number from MSB to LSB\n```\nclass Solution:\n def getDecimalValue(self, head: ListNode) -> int:\n answer = 0\n while head: \n answer = 2*answer + head.val \n head = head.next \n return answer \n``` | 244 | Given `head` which is a reference node to a singly-linked list. The value of each node in the linked list is either `0` or `1`. The linked list holds the binary representation of a number.
Return the _decimal value_ of the number in the linked list.
The **most significant bit** is at the head of the linked list.
**E... | Use dynamic programming. The state would be the index in arr1 and the index of the previous element in arr2 after sorting it and removing duplicates. |
[Python] Simple. 20ms. | convert-binary-number-in-a-linked-list-to-integer | 0 | 1 | Read the binary number from MSB to LSB\n```\nclass Solution:\n def getDecimalValue(self, head: ListNode) -> int:\n answer = 0\n while head: \n answer = 2*answer + head.val \n head = head.next \n return answer \n``` | 244 | You have a `grid` of size `n x 3` and you want to paint each cell of the grid with exactly one of the three colors: **Red**, **Yellow,** or **Green** while making sure that no two adjacent cells have the same color (i.e., no two cells that share vertical or horizontal sides have the same color).
Given `n` the number o... | Traverse the linked list and store all values in a string or array. convert the values obtained to decimal value. You can solve the problem in O(1) memory using bits operation. use shift left operation ( << ) and or operation ( | ) to get the decimal value in one operation. |
simple python solution. | convert-binary-number-in-a-linked-list-to-integer | 0 | 1 | \n# Approach\n<!-- Describe your approach to solving the problem. -->\nfirst find the length of the linked list then compute the number using loop\n\n# Complexity\n- Time complexity: $$O(N)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g... | 2 | Given `head` which is a reference node to a singly-linked list. The value of each node in the linked list is either `0` or `1`. The linked list holds the binary representation of a number.
Return the _decimal value_ of the number in the linked list.
The **most significant bit** is at the head of the linked list.
**E... | Use dynamic programming. The state would be the index in arr1 and the index of the previous element in arr2 after sorting it and removing duplicates. |
simple python solution. | convert-binary-number-in-a-linked-list-to-integer | 0 | 1 | \n# Approach\n<!-- Describe your approach to solving the problem. -->\nfirst find the length of the linked list then compute the number using loop\n\n# Complexity\n- Time complexity: $$O(N)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g... | 2 | You have a `grid` of size `n x 3` and you want to paint each cell of the grid with exactly one of the three colors: **Red**, **Yellow,** or **Green** while making sure that no two adjacent cells have the same color (i.e., no two cells that share vertical or horizontal sides have the same color).
Given `n` the number o... | Traverse the linked list and store all values in a string or array. convert the values obtained to decimal value. You can solve the problem in O(1) memory using bits operation. use shift left operation ( << ) and or operation ( | ) to get the decimal value in one operation. |
Simple Brute Force Approach | sequential-digits | 0 | 1 | # Intuition\nThe constraints make us think that thi sproblem should be solved with the help of converting the low and high to stirngs.\n\n# Approach\nCalculate the length of low and high and then run a loop from lowest length to greatest length.\nfor each length there can be only certain possibilites \nFor ex.- For len... | 1 | An integer has _sequential digits_ if and only if each digit in the number is one more than the previous digit.
Return a **sorted** list of all the integers in the range `[low, high]` inclusive that have sequential digits.
**Example 1:**
**Input:** low = 100, high = 300
**Output:** \[123,234\]
**Example 2:**
**Inp... | null |
Easy to Solve By python | sequential-digits | 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. $$O(n)$$ --... | 1 | An integer has _sequential digits_ if and only if each digit in the number is one more than the previous digit.
Return a **sorted** list of all the integers in the range `[low, high]` inclusive that have sequential digits.
**Example 1:**
**Input:** low = 100, high = 300
**Output:** \[123,234\]
**Example 2:**
**Inp... | null |
Python - math oriented | sequential-digits | 0 | 1 | # Intuition\n\nI didn\'t even consider using strings to solve this :). Oops.\n\nStarted by thinking of the properties we can leverage:\n- we can group the answers into # of digits (2, 3, etc)\n- we have a fixed initial value for each number of digits (12, 123, 1234 etc)\n- we have a fixed increment for each number of d... | 0 | An integer has _sequential digits_ if and only if each digit in the number is one more than the previous digit.
Return a **sorted** list of all the integers in the range `[low, high]` inclusive that have sequential digits.
**Example 1:**
**Input:** low = 100, high = 300
**Output:** \[123,234\]
**Example 2:**
**Inp... | null |
✔️ [Python3] ( ͡- ͜ʖ ͡-) ( ͡° ͜ʖ ͡°) ( ͡ʘ ͜ʖ ͡ʘ) LOG10, Explained | sequential-digits | 0 | 1 | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nThe main observation here is that all numbers in the result can be formed from the simple tuple `(1, 2, 3, 4, 5, 6, 7, 8, 9)` using a sliding window. What windows we should use? The range of window widths can be deri... | 18 | An integer has _sequential digits_ if and only if each digit in the number is one more than the previous digit.
Return a **sorted** list of all the integers in the range `[low, high]` inclusive that have sequential digits.
**Example 1:**
**Input:** low = 100, high = 300
**Output:** \[123,234\]
**Example 2:**
**Inp... | null |
BFS solution in Python, Time: O(1), Space: O(1) | sequential-digits | 0 | 1 | start from 1 ~ 9, and then increase by multiplying by 10 and adding last_num + 1. (last_num = num % 10)\n\nIn this solution, you don\'t even need to sort at the end.\n\n**Time: O(1)**, since you need to check at most `81` values. `(Start num: 1 ~ 9) * (At most 9 digits) = 81`\n**Space: O(1)**, since you store at most `... | 16 | An integer has _sequential digits_ if and only if each digit in the number is one more than the previous digit.
Return a **sorted** list of all the integers in the range `[low, high]` inclusive that have sequential digits.
**Example 1:**
**Input:** low = 100, high = 300
**Output:** \[123,234\]
**Example 2:**
**Inp... | null |
Python Super_Super easy #Simplest question 100% faster | sequential-digits | 0 | 1 | ```\n#(please upvote if you like the solution)\nclass Solution:\n def sequentialDigits(self, low: int, high: int) -> List[int]:\n s=\'123456789\'\n ans=[]\n for i in range(len(s)):\n for j in range(i+1,len(s)):\n st=int(s[i:j+1])\n if(st>=low and st<=high... | 29 | An integer has _sequential digits_ if and only if each digit in the number is one more than the previous digit.
Return a **sorted** list of all the integers in the range `[low, high]` inclusive that have sequential digits.
**Example 1:**
**Input:** low = 100, high = 300
**Output:** \[123,234\]
**Example 2:**
**Inp... | null |
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