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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[Python3] DP + Greedy approach + explanation | reducing-dishes | 0 | 1 | # Intuition\nWe need to maximaze the satisfaction usign formulae:\n```\nsum((i+1) * dish[i])\n```\nwith some of the dishes reduced. \nWe can have negative satisfaction dishes, which reduce the overall satisfaction, apart from they add time so that further positive-satisfaction dishes contribute more satisfaction. \n\nA... | 1 | We have a wooden plank of the length `n` **units**. Some ants are walking on the plank, each ant moves with a speed of **1 unit per second**. Some of the ants move to the **left**, the other move to the **right**.
When two ants moving in two **different** directions meet at some point, they change their directions and... | Use dynamic programming to find the optimal solution by saving the previous best like-time coefficient and its corresponding element sum. If adding the current element to the previous best like-time coefficient and its corresponding element sum would increase the best like-time coefficient, then go ahead and add it. Ot... |
easy greedy solution without using DP | reducing-dishes | 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. $$O(n)$$ -->\nO(nlogn)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 1 | A chef has collected data on the `satisfaction` level of his `n` dishes. Chef can cook any dish in 1 unit of time.
**Like-time coefficient** of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. `time[i] * satisfaction[i]`.
Return _the maximum sum... | Create an additive table that counts the sum of elements of submatrix with the superior corner at (0,0). Loop over all subsquares in O(n^3) and check if the sum make the whole array to be ones, if it checks then add 1 to the answer. |
easy greedy solution without using DP | reducing-dishes | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(nlogn)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 1 | We have a wooden plank of the length `n` **units**. Some ants are walking on the plank, each ant moves with a speed of **1 unit per second**. Some of the ants move to the **left**, the other move to the **right**.
When two ants moving in two **different** directions meet at some point, they change their directions and... | Use dynamic programming to find the optimal solution by saving the previous best like-time coefficient and its corresponding element sum. If adding the current element to the previous best like-time coefficient and its corresponding element sum would increase the best like-time coefficient, then go ahead and add it. Ot... |
5 LINES || EASY SOLUTION USING || PREFIX SUM || PYTHON | reducing-dishes | 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 | 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. |
5 LINES || EASY SOLUTION USING || PREFIX SUM || PYTHON | reducing-dishes | 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 | We have a wooden plank of the length `n` **units**. Some ants are walking on the plank, each ant moves with a speed of **1 unit per second**. Some of the ants move to the **left**, the other move to the **right**.
When two ants moving in two **different** directions meet at some point, they change their directions and... | Use dynamic programming to find the optimal solution by saving the previous best like-time coefficient and its corresponding element sum. If adding the current element to the previous best like-time coefficient and its corresponding element sum would increase the best like-time coefficient, then go ahead and add it. Ot... |
😍 C & Python Solution | detailed explained | reducing-dishes | 0 | 1 | # \uD83C\uDF40 Intuition\n## Sort the array in increasing (non-decreasing) order \uD83C\uDF4C\n- This is the base structure for maximum sum of ***like-time coefficient***\n## First, let have some mind experiment \uD83C\uDF4C\n- We know the lefter the index, the bigger the coefficient is...\n- Let\'s consider `[-9,-8,-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. |
😍 C & Python Solution | detailed explained | reducing-dishes | 0 | 1 | # \uD83C\uDF40 Intuition\n## Sort the array in increasing (non-decreasing) order \uD83C\uDF4C\n- This is the base structure for maximum sum of ***like-time coefficient***\n## First, let have some mind experiment \uD83C\uDF4C\n- We know the lefter the index, the bigger the coefficient is...\n- Let\'s consider `[-9,-8,-1... | 1 | We have a wooden plank of the length `n` **units**. Some ants are walking on the plank, each ant moves with a speed of **1 unit per second**. Some of the ants move to the **left**, the other move to the **right**.
When two ants moving in two **different** directions meet at some point, they change their directions and... | Use dynamic programming to find the optimal solution by saving the previous best like-time coefficient and its corresponding element sum. If adding the current element to the previous best like-time coefficient and its corresponding element sum would increase the best like-time coefficient, then go ahead and add it. Ot... |
easy python solution | reducing-dishes | 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. $$O(n)$$ -->o(nlogn)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 1 | A chef has collected data on the `satisfaction` level of his `n` dishes. Chef can cook any dish in 1 unit of time.
**Like-time coefficient** of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. `time[i] * satisfaction[i]`.
Return _the maximum sum... | Create an additive table that counts the sum of elements of submatrix with the superior corner at (0,0). Loop over all subsquares in O(n^3) and check if the sum make the whole array to be ones, if it checks then add 1 to the answer. |
easy python solution | reducing-dishes | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->o(nlogn)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 1 | We have a wooden plank of the length `n` **units**. Some ants are walking on the plank, each ant moves with a speed of **1 unit per second**. Some of the ants move to the **left**, the other move to the **right**.
When two ants moving in two **different** directions meet at some point, they change their directions and... | Use dynamic programming to find the optimal solution by saving the previous best like-time coefficient and its corresponding element sum. If adding the current element to the previous best like-time coefficient and its corresponding element sum would increase the best like-time coefficient, then go ahead and add it. Ot... |
Beginner-friendly || Simple solution with Greedy in Python3/TypeScript | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Intuition\nThe problem description is the following:\n- there\'s a list of `nums`\n- our goal is to extract subsequence of `nums`, whose sum is **greater**, than a rest sum\n- if there\'re multiple solutions, return the subsequence with the **minimum size** and with **maximum total sum**\n\n# Approach\n1. sort `nums`... | 1 | 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. |
Beginner-friendly || Simple solution with Greedy in Python3/TypeScript | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Intuition\nThe problem description is the following:\n- there\'s a list of `nums`\n- our goal is to extract subsequence of `nums`, whose sum is **greater**, than a rest sum\n- if there\'re multiple solutions, return the subsequence with the **minimum size** and with **maximum total sum**\n\n# Approach\n1. sort `nums`... | 1 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
Python Solution | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Intuition \n JUET\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n EASY UNDERSTANDING \n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n- o(NlogN)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n-... | 1 | 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 Solution | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Intuition \n JUET\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n EASY UNDERSTANDING \n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n- o(NlogN)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n-... | 1 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
Easy Python3 Codes | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Code1\n```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n nums=sorted(nums)[::-1]\n sum1,sum2=0,sum(nums)\n for i in range(len(nums)):\n sum1+=nums[i]\n sum2-=nums[i] \n if sum1>sum2:\n return nums[0:i+1]\... | 3 | 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. |
Easy Python3 Codes | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Code1\n```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n nums=sorted(nums)[::-1]\n sum1,sum2=0,sum(nums)\n for i in range(len(nums)):\n sum1+=nums[i]\n sum2-=nums[i] \n if sum1>sum2:\n return nums[0:i+1]\... | 3 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
Solution | minimum-subsequence-in-non-increasing-order | 0 | 1 | ```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n if (len(nums) == 1):\n return nums\n nums.sort()\n count = 0\n num = []\n l = len(nums)\n for i in range(1,l+1):\n count += nums[-i]\n num.append(nums[-i])\n ... | 2 | 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. |
Solution | minimum-subsequence-in-non-increasing-order | 0 | 1 | ```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n if (len(nums) == 1):\n return nums\n nums.sort()\n count = 0\n num = []\n l = len(nums)\n for i in range(1,l+1):\n count += nums[-i]\n num.append(nums[-i])\n ... | 2 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
Python || O(n) Solution || Easily understandable | minimum-subsequence-in-non-increasing-order | 0 | 1 | This solution might not be the fastest but the code is simple enough to understand the underlying concept. Further optimization can be done to make it faster.\n\n```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n \n nums.sort()\n totalSum=sum(nums)\n currSum=... | 1 | 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 || O(n) Solution || Easily understandable | minimum-subsequence-in-non-increasing-order | 0 | 1 | This solution might not be the fastest but the code is simple enough to understand the underlying concept. Further optimization can be done to make it faster.\n\n```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n \n nums.sort()\n totalSum=sum(nums)\n currSum=... | 1 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
Python3 simple solution | minimum-subsequence-in-non-increasing-order | 0 | 1 | ```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n nums.sort()\n l = []\n while sum(l) <= sum(nums):\n l.append(nums.pop())\n return l\n```\nIf you like the solution, please vote for this. | 12 | 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. |
Python3 simple solution | minimum-subsequence-in-non-increasing-order | 0 | 1 | ```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n nums.sort()\n l = []\n while sum(l) <= sum(nums):\n l.append(nums.pop())\n return l\n```\nIf you like the solution, please vote for this. | 12 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
Python solution | runtime - 40ms beats 92% | minimum-subsequence-in-non-increasing-order | 0 | 1 | \n# Code\n```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n total = sum(nums)\n nums.sort(reverse = True)\n sub_sum, sub_seq = 0, []\n for x in nums:\n sub_sum += x\n total -= x\n sub_seq.append(x)\n if sub_sum > t... | 3 | 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 solution | runtime - 40ms beats 92% | minimum-subsequence-in-non-increasing-order | 0 | 1 | \n# Code\n```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n total = sum(nums)\n nums.sort(reverse = True)\n sub_sum, sub_seq = 0, []\n for x in nums:\n sub_sum += x\n total -= x\n sub_seq.append(x)\n if sub_sum > t... | 3 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
#PYTHON Code using sort | minimum-subsequence-in-non-increasing-order | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n l=[]\n nums.sort(reverse=True)\n for i in range(len(nums)):\n if(sum(nums[:i+1])>(sum(nums)-sum(nums[:i+1]))):\n return nums[:i+1]\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 Code using sort | minimum-subsequence-in-non-increasing-order | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def minSubsequence(self, nums: List[int]) -> List[int]:\n l=[]\n nums.sort(reverse=True)\n for i in range(len(nums)):\n if(sum(nums[:i+1])>(sum(nums)-sum(nums[:i+1]))):\n return nums[:i+1]\n``` | 0 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
Python solution | runtime - 47 ms beats - 97.65% 💯 | minimum-subsequence-in-non-increasing-order | 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 solution | runtime - 47 ms beats - 97.65% 💯 | minimum-subsequence-in-non-increasing-order | 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 tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
python made easy | beats 88% in memory | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Intuition\nsort the give list by descending\n\n# Approach\nsort the give list nums and then lopp through nums and pop and append the first element of nums to new list arr and check on each append whether sum(arr)>sum(nums) if yes then return the arr\n\n# Complexity\n- Time complexity:\nn\n\n- Space complexity:\n1\n\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 made easy | beats 88% in memory | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Intuition\nsort the give list by descending\n\n# Approach\nsort the give list nums and then lopp through nums and pop and append the first element of nums to new list arr and check on each append whether sum(arr)>sum(nums) if yes then return the arr\n\n# Complexity\n- Time complexity:\nn\n\n- Space complexity:\n1\n\n... | 0 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
Simple Python Code | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nBy taking the sum of highest elements iteratively, we can minimize the loop iterations.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTo implement the above intuition, we will first sort the `nums` on `reverse` o... | 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. |
Simple Python Code | minimum-subsequence-in-non-increasing-order | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nBy taking the sum of highest elements iteratively, we can minimize the loop iterations.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTo implement the above intuition, we will first sort the `nums` on `reverse` o... | 0 | You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of `n` nodes numbered from `0` to `n - 1` and exactly `n - 1` `edges`. The **root** of the tree is the node `0`, and each node of the tree has **a label** which is a lower-case character given in the string `labels` (i.e. The node w... | Sort elements and take each element from the largest until accomplish the conditions. |
✔ Python3 Solution | One Line | number-of-steps-to-reduce-a-number-in-binary-representation-to-one | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(1)$$\n\n# Code\n```\nclass Solution:\n def numSteps(self, s):\n return len(s) + s.rstrip(\'0\').count(\'0\') + 2 * (s.count(\'1\') != 1) - 1\n``` | 1 | Given the binary representation of an integer as a string `s`, return _the number of steps to reduce it to_ `1` _under the following rules_:
* If the current number is even, you have to divide it by `2`.
* If the current number is odd, you have to add `1` to it.
It is guaranteed that you can always reac... | null |
✔ Python3 Solution | One Line | number-of-steps-to-reduce-a-number-in-binary-representation-to-one | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(1)$$\n\n# Code\n```\nclass Solution:\n def numSteps(self, s):\n return len(s) + s.rstrip(\'0\').count(\'0\') + 2 * (s.count(\'1\') != 1) - 1\n``` | 1 | Given a string `s` of lowercase letters, you need to find the maximum number of **non-empty** substrings of `s` that meet the following conditions:
1. The substrings do not overlap, that is for any two substrings `s[i..j]` and `s[x..y]`, either `j < x` or `i > y` is true.
2. A substring that contains a certain chara... | Read the string from right to left, if the string ends in '0' then the number is even otherwise it is odd. Simulate the steps described in the binary string. |
Why my Python solution failed on: "1111011110000011100000110001011011110010111001010111110001"? | number-of-steps-to-reduce-a-number-in-binary-representation-to-one | 0 | 1 | ```\nres = 0\nnum = int(s,2)\nwhile num != 1:\n\tif num % 2 == 0:\n\t\tnum = num / 2\n\telse:\n\t\tnum = num + 1\n\tres +=1\n\nreturn res\n```\n:-( | 8 | Given the binary representation of an integer as a string `s`, return _the number of steps to reduce it to_ `1` _under the following rules_:
* If the current number is even, you have to divide it by `2`.
* If the current number is odd, you have to add `1` to it.
It is guaranteed that you can always reac... | null |
Why my Python solution failed on: "1111011110000011100000110001011011110010111001010111110001"? | number-of-steps-to-reduce-a-number-in-binary-representation-to-one | 0 | 1 | ```\nres = 0\nnum = int(s,2)\nwhile num != 1:\n\tif num % 2 == 0:\n\t\tnum = num / 2\n\telse:\n\t\tnum = num + 1\n\tres +=1\n\nreturn res\n```\n:-( | 8 | Given a string `s` of lowercase letters, you need to find the maximum number of **non-empty** substrings of `s` that meet the following conditions:
1. The substrings do not overlap, that is for any two substrings `s[i..j]` and `s[x..y]`, either `j < x` or `i > y` is true.
2. A substring that contains a certain chara... | Read the string from right to left, if the string ends in '0' then the number is even otherwise it is odd. Simulate the steps described in the binary string. |
linear time constant space | longest-happy-string | 0 | 1 | \n# Code\n```\nclass Solution:\n def longestDiverseString(self, a: int, b: int, c: int) -> str:\n res,maxHeap = "",[]\n for count,char in [(-a,"a"),(-b,"b"),(-c,"c")]:\n if count != 0:\n heapq.heappush(maxHeap,(count,char))\n\n while maxHeap:\n count,char = h... | 1 | A string `s` is called **happy** if it satisfies the following conditions:
* `s` only contains the letters `'a'`, `'b'`, and `'c'`.
* `s` does not contain any of `"aaa "`, `"bbb "`, or `"ccc "` as a substring.
* `s` contains **at most** `a` occurrences of the letter `'a'`.
* `s` contains **at most** `b` occurr... | null |
linear time constant space | longest-happy-string | 0 | 1 | \n# Code\n```\nclass Solution:\n def longestDiverseString(self, a: int, b: int, c: int) -> str:\n res,maxHeap = "",[]\n for count,char in [(-a,"a"),(-b,"b"),(-c,"c")]:\n if count != 0:\n heapq.heappush(maxHeap,(count,char))\n\n while maxHeap:\n count,char = h... | 1 | Given an integer `n`, return **any** array containing `n` **unique** integers such that they add up to `0`.
**Example 1:**
**Input:** n = 5
**Output:** \[-7,-1,1,3,4\]
**Explanation:** These arrays also are accepted \[-5,-1,1,2,3\] , \[-3,-1,2,-2,4\].
**Example 2:**
**Input:** n = 3
**Output:** \[-1,0,1\]
**Exampl... | Use a greedy approach. Use the letter with the maximum current limit that can be added without breaking the condition. |
Python3 Solution | stone-game-iii | 0 | 1 | \n```\nclass Solution:\n def stoneGameIII(self, stoneValue: List[int]) -> str:\n n=len(stoneValue)\n dp=[0]*(n+1)\n i=n-1\n \n while i>=0:\n ans=-1001\n ans=max(ans,stoneValue[i]-dp[i+1])\n \n if i+1<n:\n ans=max(ans,stoneV... | 2 | 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 Elegant & Short | O(n) | Top-Down DP | stone-game-iii | 0 | 1 | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```\nclass Solution:\n def stoneGameIII(self, stone_value: List[int]) -> str:\n @cache\n def dp(i: int) -> int:\n take_one = stone_value[i] - dp(i + 1) \\\n if i <= n - 1 else 0\n tak... | 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 sol using Relative scores!! | stone-game-iii | 0 | 1 | # Code\n```\nclass Solution:\n def stoneGameIII(self, stoneValue: List[int]) -> str:\n n=len(stoneValue)\n\n @cache\n def dfs(idx):\n if idx>=n:\n return 0\n best=-10**20\n s=0\n for i in range(3):\n if (i+idx)>=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. |
simple python code, beats 92.38% runtime. | stone-game-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. $$O(n)$$ --... | 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. |
The evolution of solutions from naive to perfect one 🦊 🚀 | stone-game-iii | 0 | 1 | # Solution 1: Naive one\n\n## Approach\n<!-- Describe your approach to solving the problem. -->\n- Use memoization with `@cache` in function `dp(l)`.\n- Drop usage of `if - else` with `min()` and `max()` and array.\n- `-1001` is used as the minimal value according to problem constrains.\n\n## Complexity\n- Time complex... | 3 | 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. |
Beating 80% Python Easy Solution | stone-game-iii | 0 | 1 | \n\n# Code\n```\nclass Solution(object):\n def stoneGameIII(self, stoneValue):\n """\n :type stoneValue: List[int]\n :rtype: str\n """\n \n dp = [0 for _ in range(len(stoneValue))]\n if len(dp) >= 1:\n dp[-1] = stoneValue[-1]\n if len(dp) >= 2:\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 short and clean. Functional programming. | stone-game-iii | 0 | 1 | # Approach: Recursive DP\n\n# Complexity\n- Time complexity: $$O(n)$$\n\n- Space complexity: $$O(n)$$\n\nwhere, `n is number of values.`\n\n# Code\n```python\nclass Solution:\n def stoneGameIII(self, values: list[int]) -> str:\n @cache\n def score(i: int) -> int:\n return (i < len(values)) a... | 2 | 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. |
Image Explanation🏆- [Recursion->Memo->Bottom Up-> O(1) Space] - C++/Java/Python | stone-game-iii | 1 | 1 | # Video Solution (`Aryan Mittal`) - Link in LeetCode Profile\n`Stone Game III` by `Aryan Mittal`\n\n\n\n# Approach & Intution\n to compute the last digit. Generalise modulus operator idea to compute all digits. |
Python🔥Java🔥C++🔥Simple Solution🔥Easy to Understand | stone-game-iii | 1 | 1 | **!! BIG ANNOUNCEMENT !!**\nI am currently Giving away my premium content well-structured assignments and study materials to clear interviews at top companies related to computer science and data science to my current Subscribers. This is only for first 10,000 Subscribers. **DON\'T FORGET** to Subscribe\n\n# Search \u... | 19 | 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. |
Python3 Simple Code | string-matching-in-an-array | 0 | 1 | ```\narr = \' \'.join(words)\nsubStr = [i for i in words if arr.count(i) >= 2]\n \nreturn subStr\n\t\t | 95 | 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. |
Python3 Simple Code | string-matching-in-an-array | 0 | 1 | ```\narr = \' \'.join(words)\nsubStr = [i for i in words if arr.count(i) >= 2]\n \nreturn subStr\n\t\t | 95 | Given an array of integers `arr`, return _the number of subarrays with an **odd** sum_.
Since the answer can be very large, return it modulo `109 + 7`.
**Example 1:**
**Input:** arr = \[1,3,5\]
**Output:** 4
**Explanation:** All subarrays are \[\[1\],\[1,3\],\[1,3,5\],\[3\],\[3,5\],\[5\]\]
All sub-arrays sum are \[1... | Bruteforce to find if one string is substring of another or use KMP algorithm. |
Simple Explained Solution🙂 | string-matching-in-an-array | 0 | 1 | # Approach\n 1. Initialize an empty list `ans` to store the **result**.\n 2. Iterate through the `words` list using a for loop along with their indices.\n 3. For each word, concatenate all other words in the list (except the current word) into a single string `other_words`.\n 4. Check if the current word is a substring... | 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. |
Simple Explained Solution🙂 | string-matching-in-an-array | 0 | 1 | # Approach\n 1. Initialize an empty list `ans` to store the **result**.\n 2. Iterate through the `words` list using a for loop along with their indices.\n 3. For each word, concatenate all other words in the list (except the current word) into a single string `other_words`.\n 4. Check if the current word is a substring... | 4 | Given an array of integers `arr`, return _the number of subarrays with an **odd** sum_.
Since the answer can be very large, return it modulo `109 + 7`.
**Example 1:**
**Input:** arr = \[1,3,5\]
**Output:** 4
**Explanation:** All subarrays are \[\[1\],\[1,3\],\[1,3,5\],\[3\],\[3,5\],\[5\]\]
All sub-arrays sum are \[1... | Bruteforce to find if one string is substring of another or use KMP algorithm. |
Clean Python 3, suffix trie O(NlogN + N * S^2) | string-matching-in-an-array | 0 | 1 | Sort `words` with length first.\nThen add all suffix for each word to the trie after checking its existence in it.\n\nTime: `O(NlogN + N * S^2)`, where `S` is the max length of all words.\nNlogN for sorting and N * S^2 for build suffix trie.\n\nSpace: `O(N * S^2)` for suffix trie\n```\nclass Solution:\n def stringMa... | 52 | 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. |
Clean Python 3, suffix trie O(NlogN + N * S^2) | string-matching-in-an-array | 0 | 1 | Sort `words` with length first.\nThen add all suffix for each word to the trie after checking its existence in it.\n\nTime: `O(NlogN + N * S^2)`, where `S` is the max length of all words.\nNlogN for sorting and N * S^2 for build suffix trie.\n\nSpace: `O(N * S^2)` for suffix trie\n```\nclass Solution:\n def stringMa... | 52 | Given an array of integers `arr`, return _the number of subarrays with an **odd** sum_.
Since the answer can be very large, return it modulo `109 + 7`.
**Example 1:**
**Input:** arr = \[1,3,5\]
**Output:** 4
**Explanation:** All subarrays are \[\[1\],\[1,3\],\[1,3,5\],\[3\],\[3,5\],\[5\]\]
All sub-arrays sum are \[1... | Bruteforce to find if one string is substring of another or use KMP algorithm. |
Python One Liner, Improved | Pre-computed Merged Sentence | string-matching-in-an-array | 0 | 1 | Pre-computed merged sentence (a) for faster iterations.\n```\nclass Solution:\n def stringMatching(self, words: List[str]) -> List[str]:\n a = " ".join(words)\n return [w for w in words if a.count(w)>1]\n``` | 20 | 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 One Liner, Improved | Pre-computed Merged Sentence | string-matching-in-an-array | 0 | 1 | Pre-computed merged sentence (a) for faster iterations.\n```\nclass Solution:\n def stringMatching(self, words: List[str]) -> List[str]:\n a = " ".join(words)\n return [w for w in words if a.count(w)>1]\n``` | 20 | Given an array of integers `arr`, return _the number of subarrays with an **odd** sum_.
Since the answer can be very large, return it modulo `109 + 7`.
**Example 1:**
**Input:** arr = \[1,3,5\]
**Output:** 4
**Explanation:** All subarrays are \[\[1\],\[1,3\],\[1,3,5\],\[3\],\[3,5\],\[5\]\]
All sub-arrays sum are \[1... | Bruteforce to find if one string is substring of another or use KMP algorithm. |
use permutation | string-matching-in-an-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. $$O(n)$$ --... | 0 | 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. |
use permutation | string-matching-in-an-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. $$O(n)$$ --... | 0 | Given an array of integers `arr`, return _the number of subarrays with an **odd** sum_.
Since the answer can be very large, return it modulo `109 + 7`.
**Example 1:**
**Input:** arr = \[1,3,5\]
**Output:** 4
**Explanation:** All subarrays are \[\[1\],\[1,3\],\[1,3,5\],\[3\],\[3,5\],\[5\]\]
All sub-arrays sum are \[1... | Bruteforce to find if one string is substring of another or use KMP algorithm. |
Fenwick Tree | queries-on-a-permutation-with-key | 0 | 1 | # Intuition\nFenwick tree is also called BIT(binary index tree). Fenwick tree is good for getting prefixSum. How prefix sum helps with our problem?\nOur problem boils down to tracking the position of elements in the permutation. If we initilize the Fenwick tree leaves with 1 the prefix sum because the position in the p... | 1 | 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)). |
Fenwick Tree | queries-on-a-permutation-with-key | 0 | 1 | # Intuition\nFenwick tree is also called BIT(binary index tree). Fenwick tree is good for getting prefixSum. How prefix sum helps with our problem?\nOur problem boils down to tracking the position of elements in the permutation. If we initilize the Fenwick tree leaves with 1 the prefix sum because the position in the p... | 1 | You are given a string `s`.
A split is called **good** if you can split `s` into two non-empty strings `sleft` and `sright` where their concatenation is equal to `s` (i.e., `sleft + sright = s`) and the number of distinct letters in `sleft` and `sright` is the same.
Return _the number of **good splits** you can make ... | Create the permutation P=[1,2,...,m], it could be a list for example. For each i, find the position of queries[i] with a simple scan over P and then move this to the beginning. |
BRUTE FORCE || PYTHON || EASY UNDERSTANDING | queries-on-a-permutation-with-key | 0 | 1 | **90% FASTER || PYTHON || BEGINNER-FRIENDLY**\n*Upvote If you like the Solution*\n```\n ans=[]\n perm=[]\n \n for x in range(1,m+1):\n perm.append(x)\n for i in range(len(queries)):\n v=perm.index(queries[i])\n ans.append(v)\n cc=queries[i]\... | 1 | 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)). |
BRUTE FORCE || PYTHON || EASY UNDERSTANDING | queries-on-a-permutation-with-key | 0 | 1 | **90% FASTER || PYTHON || BEGINNER-FRIENDLY**\n*Upvote If you like the Solution*\n```\n ans=[]\n perm=[]\n \n for x in range(1,m+1):\n perm.append(x)\n for i in range(len(queries)):\n v=perm.index(queries[i])\n ans.append(v)\n cc=queries[i]\... | 1 | You are given a string `s`.
A split is called **good** if you can split `s` into two non-empty strings `sleft` and `sright` where their concatenation is equal to `s` (i.e., `sleft + sright = s`) and the number of distinct letters in `sleft` and `sright` is the same.
Return _the number of **good splits** you can make ... | Create the permutation P=[1,2,...,m], it could be a list for example. For each i, find the position of queries[i] with a simple scan over P and then move this to the beginning. |
python very Simple Approch easy to understand code | queries-on-a-permutation-with-key | 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 | 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 very Simple Approch easy to understand code | queries-on-a-permutation-with-key | 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 given a string `s`.
A split is called **good** if you can split `s` into two non-empty strings `sleft` and `sright` where their concatenation is equal to `s` (i.e., `sleft + sright = s`) and the number of distinct letters in `sleft` and `sright` is the same.
Return _the number of **good splits** you can make ... | Create the permutation P=[1,2,...,m], it could be a list for example. For each i, find the position of queries[i] with a simple scan over P and then move this to the beginning. |
Solution | Python | queries-on-a-permutation-with-key | 0 | 1 | ```\nclass Solution: \n def processQueries(self, queries: List[int], m: int) -> List[int]:\n # create an empty list to hold result\n result = []\n \n # create the permutation\n mList = [x for x in range(1, m + 1)]\n \n # iterate over each element from queries\n ... | 7 | 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)). |
Solution | Python | queries-on-a-permutation-with-key | 0 | 1 | ```\nclass Solution: \n def processQueries(self, queries: List[int], m: int) -> List[int]:\n # create an empty list to hold result\n result = []\n \n # create the permutation\n mList = [x for x in range(1, m + 1)]\n \n # iterate over each element from queries\n ... | 7 | You are given a string `s`.
A split is called **good** if you can split `s` into two non-empty strings `sleft` and `sright` where their concatenation is equal to `s` (i.e., `sleft + sright = s`) and the number of distinct letters in `sleft` and `sright` is the same.
Return _the number of **good splits** you can make ... | Create the permutation P=[1,2,...,m], it could be a list for example. For each i, find the position of queries[i] with a simple scan over P and then move this to the beginning. |
Python3 O(N ^ 2) solution with an array (92.46% Runtime) | queries-on-a-permutation-with-key | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nIteratively reconstruct permutation array\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Define... | 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 O(N ^ 2) solution with an array (92.46% Runtime) | queries-on-a-permutation-with-key | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nIteratively reconstruct permutation array\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Define... | 0 | You are given a string `s`.
A split is called **good** if you can split `s` into two non-empty strings `sleft` and `sright` where their concatenation is equal to `s` (i.e., `sleft + sright = s`) and the number of distinct letters in `sleft` and `sright` is the same.
Return _the number of **good splits** you can make ... | Create the permutation P=[1,2,...,m], it could be a list for example. For each i, find the position of queries[i] with a simple scan over P and then move this to the beginning. |
One-liner code Python. | html-entity-parser | 0 | 1 | # One liner Code. Python.\n# Code\n```\nclass Solution:\n def entityParser(self, text: str) -> str:\n return text.replace(\'"\', \'"\').replace(\'>\', \'>\').replace(\'<\', \'<\').replace(\''\', "\'").replace(\'&\', \'&\').replace(\'⁄\', \'/\')\n``` | 1 | **HTML entity parser** is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.
The special characters and their entities for HTML are:
* **Quotation Mark:** the entity is `"` and symbol character is `"`.
* **Single Quote Mark:** the entity ... | null |
One-liner code Python. | html-entity-parser | 0 | 1 | # One liner Code. Python.\n# Code\n```\nclass Solution:\n def entityParser(self, text: str) -> str:\n return text.replace(\'"\', \'"\').replace(\'>\', \'>\').replace(\'<\', \'<\').replace(\''\', "\'").replace(\'&\', \'&\').replace(\'⁄\', \'/\')\n``` | 1 | You are given an integer array `target`. You have an integer array `initial` of the same size as `target` with all elements initially zeros.
In one operation you can choose **any** subarray from `initial` and increment each value by one.
Return _the minimum number of operations to form a_ `target` _array from_ `initi... | Search the string for all the occurrences of the character '&'. For every '&' check if it matches an HTML entity by checking the ';' character and if entity found replace it in the answer. |
Python sol by replace and regex. 85%+ [w/ Hint ] | html-entity-parser | 0 | 1 | Python sol by replacement.\n\n---\n\n**Hint**:\n\nRemeber to put the & replacement on the last one, in order to avoid chain reaction replacement.\n\nFor example: \nshould be ![image](https://assets.leetcode.com/users/brianchiang_tw/image_158... | 13 | **HTML entity parser** is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.
The special characters and their entities for HTML are:
* **Quotation Mark:** the entity is `"` and symbol character is `"`.
* **Single Quote Mark:** the entity ... | null |
Python sol by replace and regex. 85%+ [w/ Hint ] | html-entity-parser | 0 | 1 | Python sol by replacement.\n\n---\n\n**Hint**:\n\nRemeber to put the & replacement on the last one, in order to avoid chain reaction replacement.\n\nFor example: \nshould be  | html-entity-parser | 0 | 1 | # Intuition\nJust think of string replace method.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nJust replace the all given html parsers.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n... | 1 | **HTML entity parser** is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.
The special characters and their entities for HTML are:
* **Quotation Mark:** the entity is `"` and symbol character is `"`.
* **Single Quote Mark:** the entity ... | null |
Python Simple Logic (48ms) | html-entity-parser | 0 | 1 | # Intuition\nJust think of string replace method.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nJust replace the all given html parsers.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n... | 1 | You are given an integer array `target`. You have an integer array `initial` of the same size as `target` with all elements initially zeros.
In one operation you can choose **any** subarray from `initial` and increment each value by one.
Return _the minimum number of operations to form a_ `target` _array from_ `initi... | Search the string for all the occurrences of the character '&'. For every '&' check if it matches an HTML entity by checking the ';' character and if entity found replace it in the answer. |
Python code using regex | html-entity-parser | 0 | 1 | ```\nentities = [(\'"\', \'\\"\'), (\''\', \'\\\'\'), (\'>\', \'>\'), (\'<\', \'<\'), (\'⁄\', \'/\'),(\'&\', \'&\')]\n \n for pat, repl in entities:\n\t text = re.sub(pat, repl,text)\n \n return text | 7 | **HTML entity parser** is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.
The special characters and their entities for HTML are:
* **Quotation Mark:** the entity is `"` and symbol character is `"`.
* **Single Quote Mark:** the entity ... | null |
Python code using regex | html-entity-parser | 0 | 1 | ```\nentities = [(\'"\', \'\\"\'), (\''\', \'\\\'\'), (\'>\', \'>\'), (\'<\', \'<\'), (\'⁄\', \'/\'),(\'&\', \'&\')]\n \n for pat, repl in entities:\n\t text = re.sub(pat, repl,text)\n \n return text | 7 | You are given an integer array `target`. You have an integer array `initial` of the same size as `target` with all elements initially zeros.
In one operation you can choose **any** subarray from `initial` and increment each value by one.
Return _the minimum number of operations to form a_ `target` _array from_ `initi... | Search the string for all the occurrences of the character '&'. For every '&' check if it matches an HTML entity by checking the ';' character and if entity found replace it in the answer. |
Python3 Map | html-entity-parser | 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 | **HTML entity parser** is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.
The special characters and their entities for HTML are:
* **Quotation Mark:** the entity is `"` and symbol character is `"`.
* **Single Quote Mark:** the entity ... | null |
Python3 Map | html-entity-parser | 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 an integer array `target`. You have an integer array `initial` of the same size as `target` with all elements initially zeros.
In one operation you can choose **any** subarray from `initial` and increment each value by one.
Return _the minimum number of operations to form a_ `target` _array from_ `initi... | Search the string for all the occurrences of the character '&'. For every '&' check if it matches an HTML entity by checking the ';' character and if entity found replace it in the answer. |
Describe your first thoughts on how solve problem .By using time complexity and Space complexity | html-entity-parser | 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 | **HTML entity parser** is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.
The special characters and their entities for HTML are:
* **Quotation Mark:** the entity is `"` and symbol character is `"`.
* **Single Quote Mark:** the entity ... | null |
Describe your first thoughts on how solve problem .By using time complexity and Space complexity | html-entity-parser | 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 an integer array `target`. You have an integer array `initial` of the same size as `target` with all elements initially zeros.
In one operation you can choose **any** subarray from `initial` and increment each value by one.
Return _the minimum number of operations to form a_ `target` _array from_ `initi... | Search the string for all the occurrences of the character '&'. For every '&' check if it matches an HTML entity by checking the ';' character and if entity found replace it in the answer. |
HTML Entity Parser(Runtime 45ms Beats 100.00%) | html-entity-parser | 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 | **HTML entity parser** is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.
The special characters and their entities for HTML are:
* **Quotation Mark:** the entity is `"` and symbol character is `"`.
* **Single Quote Mark:** the entity ... | null |
HTML Entity Parser(Runtime 45ms Beats 100.00%) | html-entity-parser | 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 an integer array `target`. You have an integer array `initial` of the same size as `target` with all elements initially zeros.
In one operation you can choose **any** subarray from `initial` and increment each value by one.
Return _the minimum number of operations to form a_ `target` _array from_ `initi... | Search the string for all the occurrences of the character '&'. For every '&' check if it matches an HTML entity by checking the ';' character and if entity found replace it in the answer. |
Simple Python one-liner Solution | html-entity-parser | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nJust perform replace operation\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\napply .replace() everytime\n\n# Complexity\n- Time complexity: $$O(n^2)$$ \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- ... | 0 | **HTML entity parser** is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.
The special characters and their entities for HTML are:
* **Quotation Mark:** the entity is `"` and symbol character is `"`.
* **Single Quote Mark:** the entity ... | null |
Simple Python one-liner Solution | html-entity-parser | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nJust perform replace operation\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\napply .replace() everytime\n\n# Complexity\n- Time complexity: $$O(n^2)$$ \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- ... | 0 | You are given an integer array `target`. You have an integer array `initial` of the same size as `target` with all elements initially zeros.
In one operation you can choose **any** subarray from `initial` and increment each value by one.
Return _the minimum number of operations to form a_ `target` _array from_ `initi... | Search the string for all the occurrences of the character '&'. For every '&' check if it matches an HTML entity by checking the ';' character and if entity found replace it in the answer. |
Python | Using just String Manipulation | html-entity-parser | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def entityParser(self, text: str) -> str:\n result = \'\'\n words = text.split(\'&\')\n print(words)\n for i in range(len(words)):\n if words[i].startswith(\'quot;\'):\n words[i] = \'\\"\' + words[i][5:]\n elif words[... | 0 | **HTML entity parser** is the parser that takes HTML code as input and replace all the entities of the special characters by the characters itself.
The special characters and their entities for HTML are:
* **Quotation Mark:** the entity is `"` and symbol character is `"`.
* **Single Quote Mark:** the entity ... | null |
Python | Using just String Manipulation | html-entity-parser | 0 | 1 | \n\n# Code\n```\nclass Solution:\n def entityParser(self, text: str) -> str:\n result = \'\'\n words = text.split(\'&\')\n print(words)\n for i in range(len(words)):\n if words[i].startswith(\'quot;\'):\n words[i] = \'\\"\' + words[i][5:]\n elif words[... | 0 | You are given an integer array `target`. You have an integer array `initial` of the same size as `target` with all elements initially zeros.
In one operation you can choose **any** subarray from `initial` and increment each value by one.
Return _the minimum number of operations to form a_ `target` _array from_ `initi... | Search the string for all the occurrences of the character '&'. For every '&' check if it matches an HTML entity by checking the ';' character and if entity found replace it in the answer. |
python super easy understand DP top down | number-of-ways-to-paint-n-3-grid | 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 | 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. |
dynamic programming, 32ms, beats 99.44% in Python | number-of-ways-to-paint-n-3-grid | 0 | 1 | ```\nclass Solution:\n def numOfWays(self, n: int) -> int:\n mod = 10 ** 9 + 7\n two_color, three_color = 6, 6\n for _ in range(n - 1):\n two_color, three_color = (two_color * 3 + three_color * 2) % mod, (two_color * 2 + three_color * 2) % mod\n return (two_color + three_color)... | 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. |
Math / Linear Time | number-of-ways-to-paint-n-3-grid | 0 | 1 | # Intuition\nEach of the 12 original grids (where n=1) can be divided into grids with two colors (ie: \'RGR\' which I call abas) or three colors (ie: "RGY\' aka abcs).\n\nEach aba can be followed by either 3 abas or 2 abcs, Each abc can be followed by either 2 abas or 2 abcs.\n\n# Approach\nLoop through + sum abas and ... | 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| Prefix Sum | One Pass | 99.7% beats | minimum-value-to-get-positive-step-by-step-sum | 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 of integers `nums`, you start with an initial **positive** value _startValue__._
In each iteration, you calculate the step by step sum of _startValue_ plus elements in `nums` (from left to right).
Return the minimum **positive** value of _startValue_ such that the step by step sum is never less than 1.... | Store prefix sum of all grids in another 2D array. Try all possible solutions and if you cannot find one return -1. If x is a valid answer then any y < x is also valid answer. Use binary search to find answer. |
Python| Prefix Sum | One Pass | 99.7% beats | minimum-value-to-get-positive-step-by-step-sum | 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 | You are given an undirected weighted graph of `n` nodes (0-indexed), represented by an edge list where `edges[i] = [a, b]` is an undirected edge connecting the nodes `a` and `b` with a probability of success of traversing that edge `succProb[i]`.
Given two nodes `start` and `end`, find the path with the maximum probab... | Find the minimum prefix sum. |
🔥 EASY ONE LINE SOLUTION | PYTHON | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | # Approach\nFirst of all, we have to find prefix sums. There is the great `accumulate` function from `itertools` module.\n\n```python\naccumulate(nums)\n```\n\nNext step is to find the minimum prefix, so we use `min` function.\n\n```python\nmin(accumulate(nums))\n```\n\nFor example, if the minimum is $$-4$$, then we sh... | 3 | Given an array of integers `nums`, you start with an initial **positive** value _startValue__._
In each iteration, you calculate the step by step sum of _startValue_ plus elements in `nums` (from left to right).
Return the minimum **positive** value of _startValue_ such that the step by step sum is never less than 1.... | Store prefix sum of all grids in another 2D array. Try all possible solutions and if you cannot find one return -1. If x is a valid answer then any y < x is also valid answer. Use binary search to find answer. |
🔥 EASY ONE LINE SOLUTION | PYTHON | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | # Approach\nFirst of all, we have to find prefix sums. There is the great `accumulate` function from `itertools` module.\n\n```python\naccumulate(nums)\n```\n\nNext step is to find the minimum prefix, so we use `min` function.\n\n```python\nmin(accumulate(nums))\n```\n\nFor example, if the minimum is $$-4$$, then we sh... | 3 | You are given an undirected weighted graph of `n` nodes (0-indexed), represented by an edge list where `edges[i] = [a, b]` is an undirected edge connecting the nodes `a` and `b` with a probability of success of traversing that edge `succProb[i]`.
Given two nodes `start` and `end`, find the path with the maximum probab... | Find the minimum prefix sum. |
✔Easy prefix sum solution in Python | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe maximum initial value will be 1 more than the minimum number obtained less than 1 at some step while adding the elements of array, so we have to find that non-positive number.\n\n# Approach\n<!-- Describe your approach to solving the ... | 2 | Given an array of integers `nums`, you start with an initial **positive** value _startValue__._
In each iteration, you calculate the step by step sum of _startValue_ plus elements in `nums` (from left to right).
Return the minimum **positive** value of _startValue_ such that the step by step sum is never less than 1.... | Store prefix sum of all grids in another 2D array. Try all possible solutions and if you cannot find one return -1. If x is a valid answer then any y < x is also valid answer. Use binary search to find answer. |
✔Easy prefix sum solution in Python | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe maximum initial value will be 1 more than the minimum number obtained less than 1 at some step while adding the elements of array, so we have to find that non-positive number.\n\n# Approach\n<!-- Describe your approach to solving the ... | 2 | You are given an undirected weighted graph of `n` nodes (0-indexed), represented by an edge list where `edges[i] = [a, b]` is an undirected edge connecting the nodes `a` and `b` with a probability of success of traversing that edge `succProb[i]`.
Given two nodes `start` and `end`, find the path with the maximum probab... | Find the minimum prefix sum. |
95%+ O(n) solution || Python Simple Solution | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | # Intuition\nThe smallest initial value for the sum to not dip below 1 can be found by taking a look at the lowest point the sum gets while looping through the array. Find this minimum and then add 1, since the boundery is 1 and not 0.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUse an aux a... | 2 | Given an array of integers `nums`, you start with an initial **positive** value _startValue__._
In each iteration, you calculate the step by step sum of _startValue_ plus elements in `nums` (from left to right).
Return the minimum **positive** value of _startValue_ such that the step by step sum is never less than 1.... | Store prefix sum of all grids in another 2D array. Try all possible solutions and if you cannot find one return -1. If x is a valid answer then any y < x is also valid answer. Use binary search to find answer. |
95%+ O(n) solution || Python Simple Solution | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | # Intuition\nThe smallest initial value for the sum to not dip below 1 can be found by taking a look at the lowest point the sum gets while looping through the array. Find this minimum and then add 1, since the boundery is 1 and not 0.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUse an aux a... | 2 | You are given an undirected weighted graph of `n` nodes (0-indexed), represented by an edge list where `edges[i] = [a, b]` is an undirected edge connecting the nodes `a` and `b` with a probability of success of traversing that edge `succProb[i]`.
Given two nodes `start` and `end`, find the path with the maximum probab... | Find the minimum prefix sum. |
Python Simplest Solution With Explanation | Beg to adv | Prefix sum | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | ```python\nclass Solution:\n def minStartValue(self, nums: List[int]) -> int:\n prefix_sum = 0 # taking this variable to store prefix sum\n min_start_value = 1 # as stated in the question min startValue shouldn`t be less then 1.\n \n for num in nums: # traversing through the provided list... | 6 | Given an array of integers `nums`, you start with an initial **positive** value _startValue__._
In each iteration, you calculate the step by step sum of _startValue_ plus elements in `nums` (from left to right).
Return the minimum **positive** value of _startValue_ such that the step by step sum is never less than 1.... | Store prefix sum of all grids in another 2D array. Try all possible solutions and if you cannot find one return -1. If x is a valid answer then any y < x is also valid answer. Use binary search to find answer. |
Python Simplest Solution With Explanation | Beg to adv | Prefix sum | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | ```python\nclass Solution:\n def minStartValue(self, nums: List[int]) -> int:\n prefix_sum = 0 # taking this variable to store prefix sum\n min_start_value = 1 # as stated in the question min startValue shouldn`t be less then 1.\n \n for num in nums: # traversing through the provided list... | 6 | You are given an undirected weighted graph of `n` nodes (0-indexed), represented by an edge list where `edges[i] = [a, b]` is an undirected edge connecting the nodes `a` and `b` with a probability of success of traversing that edge `succProb[i]`.
Given two nodes `start` and `end`, find the path with the maximum probab... | Find the minimum prefix sum. |
2 Lines Easy Python Solution | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | This solution is not necessary the fastest one. But it is super easy and clean (at least I think so).\n\n```\nclass Solution:\n def minStartValue(self, nums: List[int]) -> int:\n \n for i in range(1, len(nums)): nums[i] = nums[i] + nums[i - 1]\n \n return 1 if min(nums) >= 1 else abs(min(... | 5 | Given an array of integers `nums`, you start with an initial **positive** value _startValue__._
In each iteration, you calculate the step by step sum of _startValue_ plus elements in `nums` (from left to right).
Return the minimum **positive** value of _startValue_ such that the step by step sum is never less than 1.... | Store prefix sum of all grids in another 2D array. Try all possible solutions and if you cannot find one return -1. If x is a valid answer then any y < x is also valid answer. Use binary search to find answer. |
2 Lines Easy Python Solution | minimum-value-to-get-positive-step-by-step-sum | 0 | 1 | This solution is not necessary the fastest one. But it is super easy and clean (at least I think so).\n\n```\nclass Solution:\n def minStartValue(self, nums: List[int]) -> int:\n \n for i in range(1, len(nums)): nums[i] = nums[i] + nums[i - 1]\n \n return 1 if min(nums) >= 1 else abs(min(... | 5 | You are given an undirected weighted graph of `n` nodes (0-indexed), represented by an edge list where `edges[i] = [a, b]` is an undirected edge connecting the nodes `a` and `b` with a probability of success of traversing that edge `succProb[i]`.
Given two nodes `start` and `end`, find the path with the maximum probab... | Find the minimum prefix sum. |
find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | find-the-minimum-number-of-fibonacci-numbers-whose-sum-is-k | 0 | 1 | # Code\n```\nclass Solution:\n def findMinFibonacciNumbers(self, k: int) -> int:\n self.count = 0\n def fib(k):\n l = [0,1]\n for i in range(1,k+1):\n if l[-2]+l[-1]<k:\n l.append(l[-1] + l[-2])\n elif l[-2]+l[-1]==k:\n ... | 0 | Given an integer `k`, _return the minimum number of Fibonacci numbers whose sum is equal to_ `k`. The same Fibonacci number can be used multiple times.
The Fibonacci numbers are defined as:
* `F1 = 1`
* `F2 = 1`
* `Fn = Fn-1 + Fn-2` for `n > 2.`
It is guaranteed that for the given constraints we can always fin... | Use BFS. BFS on (x,y,r) x,y is coordinate, r is remain number of obstacles you can remove. |
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