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minimum-score-triangulation-of-polygon | Easy Solution With Explanation | BOTTOM UP & TOP BOTTOM | C++ | easy-solution-with-explanation-bottom-up-ahsv | [Please Upvote if it helped you ]\nint the comp function we pass the first and last index (l =0 and r = n-1)\n Now every edge of the polygon will be a side of t | Might_Guy | NORMAL | 2020-04-01T08:00:50.882782+00:00 | 2020-06-28T07:13:05.381237+00:00 | 492 | false | [Please Upvote if it helped you ]\nint the comp function we pass the first and last index (l =0 and r = n-1)\n* Now every edge of the polygon will be a side of the triangles to be made \n* lets take an edge to be made by A[0] ,A[n-1] and a vertex i in between the two\n* By taking i our problem is divided into 2 sub-pro... | 4 | 0 | [] | 2 |
minimum-score-triangulation-of-polygon | c++, bottom-up DP , easy to understand | c-bottom-up-dp-easy-to-understand-by-fig-ezlk | \nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& A) {\n int n = A.size();\n vector<vector<int>> dp(n,vector<int>(n,-1));\n | fight_club | NORMAL | 2019-11-21T13:08:09.545998+00:00 | 2019-11-21T13:08:09.546028+00:00 | 505 | false | ```\nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& A) {\n int n = A.size();\n vector<vector<int>> dp(n,vector<int>(n,-1));\n \n for(int gap = 0; gap < n; gap++){\n for(int i = 0,j=i+gap; i < n,j < n; i++,j++){\n if(gap == 0 || gap == 1){\n ... | 4 | 0 | [] | 0 |
minimum-score-triangulation-of-polygon | recursion and memorisation | recursion-and-memorisation-by-ac1-g9qd | Start with recursion hit the TLE :D\n\n def minScoreTriangulation(self, A: List[int]) -> int:\n def backtrack(start,end):\n res = sys.maxs | ac1 | NORMAL | 2019-05-18T10:10:52.451177+00:00 | 2019-05-18T10:11:46.811706+00:00 | 630 | false | Start with recursion hit the TLE :D\n```\n def minScoreTriangulation(self, A: List[int]) -> int:\n def backtrack(start,end):\n res = sys.maxsize\n if end -start +1< 3:\n return 0\n for i in range(start+1,end):\n res = min(res,backtrack(start,... | 4 | 0 | ['Backtracking', 'Memoization', 'Python'] | 1 |
minimum-score-triangulation-of-polygon | Variant of MATRIX CHAIN MULTIPLICATION | DP | DRY RUN attached | variant-of-matrix-chain-multiplication-d-3bx9 | Intuition\nThere are three variable which are used i (start point), j (end point) & k (start -> end). This shows its a variation of MATRIX CHAIN MULTIPLICATION\ | JigarSiddhpura | NORMAL | 2024-07-29T07:36:45.967458+00:00 | 2024-07-29T07:36:45.967490+00:00 | 241 | false | # Intuition\nThere are three variable which are used `i` (start point), `j` (end point) & `k` (start -> end). This shows its a variation of `MATRIX CHAIN MULTIPLICATION`\n\n# Dry Run for [3,7,4,5]\n\nclass Solution1:\n def minScoreTriangulation(self, values: List[int]) -> int:\n | pulkit_uppal | NORMAL | 2023-09-30T09:13:16.478901+00:00 | 2023-09-30T09:13:16.478925+00:00 | 610 | false | ```\n#Recursive\n#Time Complexity: Exponential\n#Space Complexity: O(n)\nclass Solution1:\n def minScoreTriangulation(self, values: List[int]) -> int:\n def solve(i, j):\n if i+1 == j:\n return 0\n m = float(\'inf\')\n for k in range(i+1, j):\n m ... | 3 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Python', 'Python3'] | 1 |
minimum-score-triangulation-of-polygon | C++ Aditya Verma's Approach ✅✅ | c-aditya-vermas-approach-by-akshay_ar_20-ltk5 | Intuition\n Describe your first thoughts on how to solve this problem. \n- Matrix Chain Multiplication [M C M]\n\n# Approach\n Describe your approach to solving | akshay_AR_2002 | NORMAL | 2023-04-08T19:16:11.638547+00:00 | 2023-04-08T19:16:11.638589+00:00 | 884 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- Matrix Chain Multiplication [M C M]\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- First initialize the dynamic programming array \'dp\' with -1. This is done to indicate that a particular subproblem has not b... | 3 | 0 | ['C++'] | 1 |
minimum-score-triangulation-of-polygon | Minimum Score Triangulation of Polygon similar to matrix chain multiplication problem | minimum-score-triangulation-of-polygon-s-xwwd | \nclass Solution {\npublic:\n int f(int i,int j,vector<int>& values,vector<vector<int>>&dp)\n {\n if(i==j) return 0;\n int mini=1e9;\n | riturajkumar7256 | NORMAL | 2022-07-29T08:13:51.596156+00:00 | 2022-10-13T14:42:55.675906+00:00 | 436 | false | ```\nclass Solution {\npublic:\n int f(int i,int j,vector<int>& values,vector<vector<int>>&dp)\n {\n if(i==j) return 0;\n int mini=1e9;\n if(dp[i][j]!=-1) return dp[i][j];\n \n for(int k=i;k<j;k++)\n {\n int steps=values[i-1]*values[k]*values[j]+f(i,k,values,dp)+... | 3 | 0 | ['Dynamic Programming', 'C'] | 1 |
minimum-score-triangulation-of-polygon | C++ Explained | MCM variation | c-explained-mcm-variation-by-bit_legion-y9cm | Because we need to check for every possible combination of sides, therefore, we can approach this question by MCM. \n\n\nclass Solution {\npublic:\n \n in | biT_Legion | NORMAL | 2021-06-15T03:58:41.602178+00:00 | 2021-06-15T03:58:41.602221+00:00 | 453 | false | Because we need to check for every possible combination of sides, therefore, we can approach this question by MCM. \n\n```\nclass Solution {\npublic:\n \n int dp[1005][1005];\n \n int MSTP(vector <int> &arr, int i, int j){\n if(i >= j)\n return 0;\n \n if(dp[i][j] != -1)\n ... | 3 | 1 | ['Dynamic Programming', 'C'] | 0 |
minimum-score-triangulation-of-polygon | Java DP, easy to understand. Just few lines | java-dp-easy-to-understand-just-few-line-e01e | Given vertices [0, n-1]\uFF0Cchoose one of them between 0 & n-1, say vertice i. \n\nThe whole polygon could be splited into 3 parts,\n1. A triangle formed by 3 | ryan7887 | NORMAL | 2021-01-17T03:12:28.462055+00:00 | 2021-01-17T03:12:28.462081+00:00 | 140 | false | Given vertices [0, n-1]\uFF0Cchoose one of them between 0 & n-1, say vertice i. \n\nThe whole polygon could be splited into 3 parts,\n1. A triangle formed by 3 vertices 0, i, n-1\n2. A polygon formed by vertices [0, i]\n3. A polygon formed by vertices [i, n-1]\n\nNow the problem is transformed into "get the minimum va... | 3 | 0 | [] | 0 |
minimum-score-triangulation-of-polygon | [PYTHON 3] DP | Iterative Solution | python-3-dp-iterative-solution-by-mohame-yf9y | \nclass Solution:\n def minScoreTriangulation(self, A: List[int]) -> int:\n n = len(A)\n dp = [[0 for i in range(n)] for j in range(n)]\n | mohamedimranps | NORMAL | 2020-06-07T15:37:37.032948+00:00 | 2020-06-07T15:37:37.033003+00:00 | 547 | false | ```\nclass Solution:\n def minScoreTriangulation(self, A: List[int]) -> int:\n n = len(A)\n dp = [[0 for i in range(n)] for j in range(n)]\n for k in range(2 , n):\n for i in range(n - k):\n start , end = i , i + k\n dp[start][end] = float("inf")\n ... | 3 | 0 | ['Dynamic Programming', 'Iterator', 'Python3'] | 1 |
minimum-score-triangulation-of-polygon | Ruby 100%. Explanation. Image. | ruby-100-explanation-image-by-user9697n-8mcd | Leetcode: 1039. Minimum Score Triangulation of Polygon.\n\nThis is a recursive function. To calculate a minimum split into triangle pices we select one edge bet | user9697n | NORMAL | 2020-04-09T17:59:10.804437+00:00 | 2020-04-09T17:59:10.804493+00:00 | 276 | false | #### Leetcode: 1039. Minimum Score Triangulation of Polygon.\n\nThis is a recursive function. To calculate a minimum split into triangle pices we select one **edge** between to vertex (let it be an edge between first and last vertex). And draw all possible triangles with this **edge**. It will be **N-2** triangles, bec... | 3 | 0 | ['Ruby'] | 0 |
minimum-score-triangulation-of-polygon | Two Solutions in Python 3 (DP) (Top Down and Bottom Up) | two-solutions-in-python-3-dp-top-down-an-ypgb | DP - Top Down - With Recursion (Slower): (seven lines)\n\nclass Solution:\n def minScoreTriangulation(self, A: List[int]) -> int:\n \tSP, LA = [[0]*50 for | junaidmansuri | NORMAL | 2019-09-27T04:25:42.333075+00:00 | 2019-09-27T05:43:46.093662+00:00 | 1,014 | false | _DP - Top Down - With Recursion (Slower):_ (seven lines)\n```\nclass Solution:\n def minScoreTriangulation(self, A: List[int]) -> int:\n \tSP, LA = [[0]*50 for i in range(50)], len(A)\n \tdef MinPoly(a,b):\n \t\tL, m = b - a + 1, math.inf; \n \t\tif SP[a][b] != 0 or L < 3: return SP[a][b]\n \t\tfor i ... | 3 | 1 | ['Dynamic Programming', 'Python', 'Python3'] | 0 |
minimum-score-triangulation-of-polygon | java dp | java-dp-by-sumonon-7qx6 | It is always the matter of modeling.\nIn this problem, the key step is that when we take out any triangle from a polygen, the remain parts of the polygen can be | sumonon | NORMAL | 2019-08-06T14:00:06.819978+00:00 | 2019-08-06T14:00:06.820014+00:00 | 183 | false | It is always the matter of modeling.\nIn this problem, the key step is that when we take out any triangle from a polygen, the remain parts of the polygen can be split up to smaller polygen but faces same kind of problems, which constructs subproblems here.\n\nfrom this point, a top-down version that is easier to under... | 3 | 0 | [] | 0 |
minimum-score-triangulation-of-polygon | Java code inspired by votrubac's solution. | java-code-inspired-by-votrubacs-solution-3xjg | This below solution is just an implementation in java of an awesome solution by votrubac here : https://leetcode.com/problems/minimum-score-triangulation-of-pol | successinvain | NORMAL | 2019-05-07T16:01:13.138001+00:00 | 2019-05-07T16:01:13.138040+00:00 | 303 | false | This below solution is just an implementation in java of an awesome solution by votrubac here : https://leetcode.com/problems/minimum-score-triangulation-of-polygon/discuss/286753/C%2B%2B-with-picture\n```\n//Algorithm:\n//pick a side with i, j vertices, pick an anchor (k) to form a triangle.\n// move the anchor (k) al... | 3 | 0 | [] | 0 |
minimum-score-triangulation-of-polygon | [Java] Memoization (Top Down) | java-memoization-top-down-by-ztztzt8888-wq9b | \n\tpublic static int minScoreTriangulation(int[] arr) {\n int len = arr.length;\n int[][] lookup = new int[len][len];\n return minScoreFro | ztztzt8888 | NORMAL | 2019-05-05T04:29:35.256556+00:00 | 2019-05-05T04:29:35.256660+00:00 | 427 | false | ```\n\tpublic static int minScoreTriangulation(int[] arr) {\n int len = arr.length;\n int[][] lookup = new int[len][len];\n return minScoreFromTo(arr, 0, len - 1, lookup);\n }\n\n private static int minScoreFromTo(int[] arr, int from, int to, int[][] lookup) {\n if (from >= to || from ... | 3 | 0 | [] | 1 |
minimum-score-triangulation-of-polygon | DP Java | dp-java-by-poorvank-n2e0 | Try all possible combinations.\n\n\nLet Minimum Cost of triangulation of vertices from i to j be min(i, j)\nIf j <= i + 2 Then\n min(i, j) = 0\nElse\n min(i, | poorvank | NORMAL | 2019-05-05T04:03:09.938442+00:00 | 2019-05-05T04:03:09.938487+00:00 | 443 | false | Try all possible combinations.\n\n```\nLet Minimum Cost of triangulation of vertices from i to j be min(i, j)\nIf j <= i + 2 Then\n min(i, j) = 0\nElse\n min(i, j) = Math.min { min(i, k) + min(k, j) + (A[i]*A[j]*A[k]) } i+1<=k<=j-1\n```\n\n```\npublic int minScoreTriangulation(int[] A) {\n int n = A.length;\n... | 3 | 1 | [] | 0 |
minimum-score-triangulation-of-polygon | Minimum Score Triangulation of Polygon | minimum-score-triangulation-of-polygon-b-1cxi | Code | Ansh1707 | NORMAL | 2025-03-27T20:12:15.123729+00:00 | 2025-03-27T20:12:15.123729+00:00 | 48 | false |
# Code
```python []
class Solution(object):
def minScoreTriangulation(self, values):
"""
:type values: List[int]
:rtype: int
"""
n = len(values)
dp = [[0] * n for _ in range(n)]
for length in range(2, n):
for i in range(n - length):
... | 2 | 0 | ['Array', 'Dynamic Programming', 'Python'] | 0 |
minimum-score-triangulation-of-polygon | Simple C++ Solution | simple-c-solution-by-divyanshu_singh_cs-r3x9 | 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 | Divyanshu_singh_cs | NORMAL | 2023-07-27T11:10:04.165512+00:00 | 2023-07-27T11:10:04.165535+00:00 | 350 | false | # 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 | 0 | ['C++'] | 0 |
minimum-score-triangulation-of-polygon | Recursive, Memoization, Tabulation Approach Java | recursive-memoization-tabulation-approac-elfk | Complexity\n- Time complexity: O(n^3) for tabulation\n\n- Space complexity: O(n^2)\n\n# Code\n\nclass Solution {\n public int minScoreTriangulation(int[] val | athravmehta06 | NORMAL | 2023-04-25T16:34:45.000220+00:00 | 2023-04-25T16:34:45.000276+00:00 | 1,238 | false | # Complexity\n- Time complexity: $$O(n^3)$$ for tabulation\n\n- Space complexity: $$O(n^2)$$\n\n# Code\n```\nclass Solution {\n public int minScoreTriangulation(int[] values) {\n int n = values.length;\n // return helperRec(values, 0, n - 1);\n\n int[][] dp = new int[n + 1][n + 1];\n for(... | 2 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Java'] | 0 |
minimum-score-triangulation-of-polygon | Minimum Score Triangulation of Polygon(Matrix Multiplication) - Java sol | minimum-score-triangulation-of-polygonma-5lt9 | \n\n# Code\n\nclass Solution {\n public int minScoreTriangulation(int[] values) {\n int N = values.length;\n int[][] dp = new int[N][N];\n | whopiyushanand | NORMAL | 2023-02-20T15:18:07.674776+00:00 | 2023-02-20T15:18:07.674819+00:00 | 1,334 | false | \n\n# Code\n```\nclass Solution {\n public int minScoreTriangulation(int[] values) {\n int N = values.length;\n int[][] dp = new int[N][N];\n for(int len=2; len<N; len++){\n for(int row=0, col=len; row<N-len; row++, col++){\n dp[row][col] = Integer.MAX_VALUE;\n ... | 2 | 0 | ['Dynamic Programming', 'Java'] | 0 |
minimum-score-triangulation-of-polygon | Matrix Chain Multiplication || DP || Memoization | matrix-chain-multiplication-dp-memoizati-w775 | 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 | _shiv_70 | NORMAL | 2023-01-11T11:22:46.154953+00:00 | 2023-01-11T11:23:20.709345+00:00 | 1,320 | false | # 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:\nO(N^3)\n\n- Space complexity:\nO(N*N)+O(N)\n\n# Code\n```\nclass Solution {\npublic:\nint fun(int i,int j,vector<int>& arr, vector<... | 2 | 0 | ['Dynamic Programming', 'Memoization', 'C++'] | 1 |
minimum-score-triangulation-of-polygon | c++ | easy | short | c-easy-short-by-venomhighs7-l6d4 | \n\n# Code\n\nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& A) {\n int n = A.size();\n vector<vector<int>> dp(n, vector<int | venomhighs7 | NORMAL | 2022-11-02T04:06:09.419134+00:00 | 2022-11-02T04:06:09.419167+00:00 | 2,143 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& A) {\n int n = A.size();\n vector<vector<int>> dp(n, vector<int>(n));\n for (int j = 2; j < n; ++j) {\n for (int i = j - 2; i >= 0; --i) {\n dp[i][j] = INT_MAX;\n for (int ... | 2 | 0 | ['C++'] | 0 |
minimum-score-triangulation-of-polygon | Java Solutions Recursion,Memoization and Bottom up DP | java-solutions-recursionmemoization-and-naatj | Simple Recusion\n\nclass Solution {\npublic int minScoreTriangulation(int[] values) {\n int n=values.length-1;\n return solve(values,0,n);\n }\ | jaswinder_97 | NORMAL | 2022-10-21T03:50:03.535557+00:00 | 2022-10-21T03:50:03.535602+00:00 | 555 | false | Simple Recusion\n```\nclass Solution {\npublic int minScoreTriangulation(int[] values) {\n int n=values.length-1;\n return solve(values,0,n);\n }\n private int solve(int[] values,int i,int j){\n if(i+1==j) return 0;\n int ans=Integer.MAX_VALUE;\n for(int k=i+1;k<j;k++){\n ... | 2 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization'] | 0 |
minimum-score-triangulation-of-polygon | DP Solution in Javascript (Recursion + Dp Memo + Dp tabulation) | dp-solution-in-javascript-recursion-dp-m-9dps | 1. Recursion\n\n\nfunction solveRec(value, i, j) {\n // base case\n if (i + 1 === j) return 0;\n let ans = Number.MAX_VALUE;\n for (let k = i + 1; k < j; k+ | vivekdogra | NORMAL | 2022-08-03T19:29:51.409852+00:00 | 2022-08-03T19:29:51.409891+00:00 | 226 | false | **1. Recursion**\n\n````\nfunction solveRec(value, i, j) {\n // base case\n if (i + 1 === j) return 0;\n let ans = Number.MAX_VALUE;\n for (let k = i + 1; k < j; k++) {\n ans = Math.min(\n ans,\n value[i] * value[j] * value[k] +\n solveRec(value, i, k) +\n solveRec(value, k, j)\n );\n ... | 2 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'JavaScript'] | 2 |
minimum-score-triangulation-of-polygon | C++ || Memoization || Tabulation | c-memoization-tabulation-by-tejasdarwai-d40f | Memoization\n\nint solve(vector<int> &values, int i, int j, vector<vector<int>> &dp){\n if(i+1==j){\n return 0;\n }\n if(dp[i][j | TejasDarwai | NORMAL | 2022-07-21T09:08:44.811899+00:00 | 2022-07-21T09:08:44.811955+00:00 | 228 | false | Memoization\n```\nint solve(vector<int> &values, int i, int j, vector<vector<int>> &dp){\n if(i+1==j){\n return 0;\n }\n if(dp[i][j]!=-1){\n return dp[i][j];\n }\n int ans=INT_MAX;\n for(int k=i+1; k<j; k++){\n ans = min(ans, (values[i]*values[j... | 2 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C'] | 0 |
minimum-score-triangulation-of-polygon | 1039. Minimum Score Triangulation of Polygon | 1039-minimum-score-triangulation-of-poly-j9pe | // This is nothing but matrix chain multiplication .\n\n\nclass Solution {\n public int minScoreTriangulation(int[] values) {\n int N = values.length; | Ardhendu_init_ | NORMAL | 2022-07-14T08:11:58.140476+00:00 | 2022-07-14T08:13:54.934841+00:00 | 424 | false | **// This is nothing but matrix chain multiplication .**\n\n```\nclass Solution {\n public int minScoreTriangulation(int[] values) {\n int N = values.length;\n int dp [][] = new int [N][N];\n for(int i = 0 ; i < N ; i++){\n for(int j = 0 ; j < N ; j++){\n dp[i][j]= -1;\... | 2 | 0 | ['Dynamic Programming', 'Memoization', 'Java'] | 0 |
minimum-score-triangulation-of-polygon | Matrix Chain Multiplication | Tabulation | matrix-chain-multiplication-tabulation-b-g471 | Same as Matrix Chain Multiplication \n\nclass Solution {\npublic:\n// Time Complexity -> O(N^3) \n// Space Complexity -> O(N^2)\n int minScoreTriangulation(v | _limitlesspragma | NORMAL | 2022-05-30T17:06:06.950631+00:00 | 2022-05-30T17:06:06.950853+00:00 | 156 | false | # ***Same as Matrix Chain Multiplication*** \n```\nclass Solution {\npublic:\n// Time Complexity -> O(N^3) \n// Space Complexity -> O(N^2)\n int minScoreTriangulation(vector<int>& arr) {\n int n=arr.size();\n \n vector<vector<int>> dp(n, vector<int>(n,0));\n \n for(int i=n-2;i>=1;i... | 2 | 0 | ['Dynamic Programming'] | 0 |
minimum-score-triangulation-of-polygon | Simple Python DFS with Explanation (12 lines) | simple-python-dfs-with-explanation-12-li-1me5 | the intuition here is that l and r are ALWAYS going to be in a triange\nwe just need to figure out the third point in the triangle\npossible third points are al | jaredlwong | NORMAL | 2022-02-15T04:35:53.269740+00:00 | 2022-02-15T04:35:53.269789+00:00 | 276 | false | the intuition here is that l and r are ALWAYS going to be in a triange\nwe just need to figure out the third point in the triangle\npossible third points are all indices between l and r\n\nonce you draw the lines between l, r and some i between l and r\nyou have two subproblems and one triangle\ntwo subproblems are l t... | 2 | 0 | ['Depth-First Search', 'Python'] | 0 |
minimum-score-triangulation-of-polygon | Aditya Verma approach - recursion memoization - JAVA - | aditya-verma-approach-recursion-memoizat-wkt7 | \nclass Solution {\n int[][] t = new int[51][51];\n public int minScoreTriangulation(int[] values) \n {\n for(int i=0;i<51;i++)\n {\n | siddharth_78 | NORMAL | 2021-08-07T05:24:11.831041+00:00 | 2022-03-31T01:52:56.876975+00:00 | 235 | false | \nclass Solution {\n int[][] t = new int[51][51];\n public int minScoreTriangulation(int[] values) \n {\n for(int i=0;i<51;i++)\n {\n for(int j=0;j<51;j++)\n {\n t[i][j] = -1;\n }\n }\n \n return solve(values,0,values.length - 1... | 2 | 5 | [] | 1 |
minimum-score-triangulation-of-polygon | C++ | dynamic programming | Using gap strategy | c-dynamic-programming-using-gap-strategy-5coj | \nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& values) {\n int n = values.size();\n vector<vector<int>> dp(n,vector<int>( | armangupta48 | NORMAL | 2021-05-02T23:08:43.751967+00:00 | 2021-05-02T23:08:43.752006+00:00 | 352 | false | ```\nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& values) {\n int n = values.size();\n vector<vector<int>> dp(n,vector<int>(n,0));\n for(int g = 0;g<n;g++)\n {\n for(int i = 0,j = g;j<n;i++,j++)\n {\n if(g==0 || g==1)\n ... | 2 | 0 | ['Dynamic Programming', 'C', 'C++'] | 1 |
minimum-score-triangulation-of-polygon | Java DP | java-dp-by-vardhamank93-vc0p | \nclass Solution {\n public int minScoreTriangulation(int[] A) {\n int[][] dp = new int[A.length][A.length];\n \n for(int g = 0; g < dp. | vardhamank93 | NORMAL | 2020-11-27T09:21:49.638301+00:00 | 2020-11-27T09:21:49.638335+00:00 | 273 | false | ```\nclass Solution {\n public int minScoreTriangulation(int[] A) {\n int[][] dp = new int[A.length][A.length];\n \n for(int g = 0; g < dp.length; g++){\n for(int i = 0,j = g; j < dp[0].length; i++,j++){\n if(g == 0){\n dp[i][j] = 0; // trivial case a... | 2 | 0 | ['Dynamic Programming', 'Java'] | 0 |
minimum-score-triangulation-of-polygon | 4 ms C++ solution beats 100% of all submissions, top down approach | 4-ms-c-solution-beats-100-of-all-submiss-2p7c | \nint dp[55][55];\nint minScore(vector<int>& A,int n,int i,int j){\n \n if(dp[i][j] != -1) return dp[i][j];\n if(j == 0) j = n-1;\n int res = INT_ | vishalnsit | NORMAL | 2020-06-07T06:06:30.480075+00:00 | 2020-06-07T06:06:30.480122+00:00 | 346 | false | ```\nint dp[55][55];\nint minScore(vector<int>& A,int n,int i,int j){\n \n if(dp[i][j] != -1) return dp[i][j];\n if(j == 0) j = n-1;\n int res = INT_MAX;\n bool loopRun = false;\n for(int k=i+1;k<j;k++){\n loopRun = true;\n int temp = (A[i]*A[j]*A[k]) + minScore(A,n,i,k) + minScore(A,... | 2 | 0 | ['Dynamic Programming', 'Memoization', 'C'] | 0 |
minimum-score-triangulation-of-polygon | C++ bottom up and memoization solution with explanation | c-bottom-up-and-memoization-solution-wit-5e2d | \n/*\n /*\n https://leetcode.com/problems/minimum-score-triangulation-of-polygon/submissions/\n \n The idea is to take each pair of vertices possibl | cryptx_ | NORMAL | 2020-01-06T06:20:47.449195+00:00 | 2020-01-06T06:40:04.621874+00:00 | 442 | false | ```\n/*\n /*\n https://leetcode.com/problems/minimum-score-triangulation-of-polygon/submissions/\n \n The idea is to take each pair of vertices possible and then with those fixed, find \n a vertex in between such that the polygon on left and right side of it are of min score.\n \n*/\n\nclass Solution ... | 2 | 0 | [] | 0 |
minimum-score-triangulation-of-polygon | c++ memorized DFS solution in O(n^3) complexity | c-memorized-dfs-solution-in-on3-complexi-928h | \nclass Solution {\npublic:\n vector<vector<int>> dp;\n int memorizedDFS(vector<int>& A, int start, int end){\n if(start + 1 == end)\n r | mintyiqingchen | NORMAL | 2019-09-11T08:29:44.244421+00:00 | 2019-10-07T15:57:00.231528+00:00 | 331 | false | ```\nclass Solution {\npublic:\n vector<vector<int>> dp;\n int memorizedDFS(vector<int>& A, int start, int end){\n if(start + 1 == end)\n return 0;\n if(dp[start][end] != -1)\n return dp[start][end];\n \n if(start + 2 == end){\n dp[start][end] = A[start... | 2 | 0 | [] | 1 |
minimum-score-triangulation-of-polygon | [Java] DP | java-dp-by-peritan-xbg1 | dp[i][j] = min cost for A[i..j]\nbase case: if j - i + 1 == 3 (length == 3), dp[i][j] = A[i]A[i+1]A[i+2]\ndp[i][j] = dp[i][k] + dp[k][j] + A[i]A[j]A[k]\n\nin bo | peritan | NORMAL | 2019-05-05T04:05:37.249794+00:00 | 2019-05-05T05:12:57.973013+00:00 | 237 | false | dp[i][j] = min cost for A[i..j]\nbase case: if j - i + 1 == 3 (length == 3), dp[i][j] = A[i]*A[i+1]*A[i+2]\ndp[i][j] = dp[i][k] + dp[k][j] + A[i]*A[j]*A[k]\n\nin bottom up approach\none key point is we should constructure the answer start from base case\nyou can draw a graph, start from side = 4, when you separate the ... | 2 | 1 | [] | 0 |
minimum-score-triangulation-of-polygon | Marvelous memoization :) | marvelous-memoization-by-pradyumnaprahas-uamp | Intuition\n Describe your first thoughts on how to solve this problem. \nFirst come up with a backtracking solution trying all possible combinations then later | PradyumnaPrahas2_2 | NORMAL | 2024-12-03T15:00:33.192661+00:00 | 2024-12-03T15:00:33.192688+00:00 | 25 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst come up with a backtracking solution trying all possible combinations then later optimize using 2d array to avoid repeatitive calculations.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nCome up with a backtra... | 1 | 0 | ['Dynamic Programming', 'Backtracking', 'Java'] | 0 |
minimum-score-triangulation-of-polygon | Best C++ Solution | 0ms, beats 100% | DP | best-c-solution-0ms-beats-100-dp-by-prat-wol9 | Code\ncpp []\nclass Solution {\npublic:\n int solve(vector<int>& v, int i, int j, vector<vector<int>>& dp) {\n if(i+1 == j) return 0;\n\n if(dp | prateek_sen | NORMAL | 2024-10-29T04:34:44.454784+00:00 | 2024-10-29T04:34:44.454809+00:00 | 26 | false | # Code\n```cpp []\nclass Solution {\npublic:\n int solve(vector<int>& v, int i, int j, vector<vector<int>>& dp) {\n if(i+1 == j) return 0;\n\n if(dp[i][j] != -1) return dp[i][j];\n\n int ans = INT_MAX;\n for(int k=i+1; k<j; k++) {\n ans = min(ans, v[i]*v[j]*v[k] + solve(v, k, j... | 1 | 0 | ['C++'] | 0 |
minimum-score-triangulation-of-polygon | ✅ One Line Solution | one-line-solution-by-mikposp-khqb | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n \n\nTime complexity: O(n^2). Space complexity: O( | MikPosp | NORMAL | 2024-03-03T11:00:34.523894+00:00 | 2024-03-03T11:00:34.523928+00:00 | 113 | false | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n<!-- -->\n\nTime complexity: $$O(n^2)$$. Space complexity: $$O(n^2)$$\n```\nclass Solution:\n def minScoreTriangulation(self, v: List[int]) -> int:\n return (f:=cache(lambda i,j:j-i>1 and min(f(i,k)+... | 1 | 0 | ['Array', 'Dynamic Programming', 'Recursion', 'Memoization', 'Python', 'Python3'] | 0 |
minimum-score-triangulation-of-polygon | Easy solution || using Recursion or Memoization or Tabulation || | easy-solution-using-recursion-or-memoiza-k1av | 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 | zephyrus17 | NORMAL | 2024-02-03T05:23:26.060221+00:00 | 2024-02-03T05:23:26.060247+00:00 | 365 | false | # 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 | 0 | ['C++'] | 0 |
minimum-score-triangulation-of-polygon | Python DP Solution, Faster than 94% | python-dp-solution-faster-than-94-by-div-7pf9 | \n Describe your first thoughts on how to solve this problem. \n\n# Approach\nWe can solve it with Dynamic programming. DP(pos1,pos2) is the minimum cost of tri | Divyanshuyyadav | NORMAL | 2023-09-29T14:00:41.254545+00:00 | 2023-09-29T14:00:41.254578+00:00 | 100 | false | \n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nWe can solve it with Dynamic programming. DP(pos1,pos2) is the minimum cost of triangulation of vertices from pos1 to pos2. if (pos2-pos1<2) return 0 means its not possible to get any triangle. Else, we do DP(pos1,pos2)= min(DP(pos1,po... | 1 | 0 | ['Python3'] | 0 |
minimum-score-triangulation-of-polygon | dynamic programming - Problem Pattern | Matrix Chain Multiplication || | dynamic-programming-problem-pattern-matr-nazc | Intuition\n Describe your first thoughts on how to solve this problem. \nThis is the child problem of MCM .\n\nProblem Description:\nThe problem involves findin | imsej_al | NORMAL | 2023-08-30T11:39:07.925249+00:00 | 2023-08-30T11:39:07.925278+00:00 | 20 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis is the child problem of MCM .\n\nProblem Description:\nThe problem involves finding the minimum score needed to triangulate a convex polygon formed by a sequence of vertices, where each vertex has an associated value. Triangulation r... | 1 | 0 | ['Array', 'Dynamic Programming', 'Memoization', 'C++'] | 0 |
minimum-score-triangulation-of-polygon | C++ || DP solution || Tabulation method | c-dp-solution-tabulation-method-by-hey_h-2ont | Intuition\nMatrix Chain Multiplication [M C M] . DP solution using tabulation method.\n\n\n\n# Complexity\n- Time complexity:O(n^3)\n Add your time complexity h | Hey_Himanshu | NORMAL | 2023-06-06T14:08:54.595550+00:00 | 2023-06-06T14:08:54.595590+00:00 | 14 | false | # Intuition\nMatrix Chain Multiplication [M C M] . DP solution using tabulation method.\n\n\n\n# Complexity\n- Time complexity:O(n^3)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n^2)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n\n# Code\n```\n#include<bits/stdc++.h>\nusi... | 1 | 0 | ['Dynamic Programming', 'Matrix', 'C++'] | 0 |
minimum-score-triangulation-of-polygon | Recursive Triangulation with Dynamic Programming | recursive-triangulation-with-dynamic-pro-mio8 | Intuition\nWe have to recursively form every possible triangles \n\n# Approach\nWe select first index and last index as a base, and the function recursively fin | harsh_reality_ | NORMAL | 2023-06-03T13:16:27.477378+00:00 | 2023-06-03T13:16:27.477423+00:00 | 172 | false | # Intuition\nWe have to recursively form every possible triangles \n\n# Approach\nWe select first index and last index as a base, and the function recursively find the best possible third index\n\n# Complexity\n-Time complexity:\nO(n^3)\n\n-Space complexity:\nO(n^2) + O(n)\n\n# Recursive and Memoized Code \n```\nclass ... | 1 | 0 | ['Divide and Conquer', 'Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
minimum-score-triangulation-of-polygon | Beats 100% | Java | Matrix Chain Multiplication | beats-100-java-matrix-chain-multiplicati-02s2 | \n# Complexity\n- Time complexity: n^2\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: n^2\n Add your space complexity here, e.g. O(n) \n\n# | PrashantNegi878 | NORMAL | 2023-04-29T07:38:18.382222+00:00 | 2023-04-29T07:46:15.572022+00:00 | 822 | false | \n# Complexity\n- Time complexity: n^2\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: n^2\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n int[][] dp;\n public int minScoreTriangulation(int[] values) {\n int l=values.length;\n ... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Matrix', 'Java'] | 1 |
minimum-score-triangulation-of-polygon | python super easy dp top down | python-super-easy-dp-top-down-by-harrych-umb9 | 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 | harrychen1995 | NORMAL | 2023-01-24T16:01:14.352164+00:00 | 2023-01-24T16:05:58.425111+00:00 | 140 | false | # 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 | 0 | ['Python3'] | 0 |
minimum-score-triangulation-of-polygon | Just a runnable solution | just-a-runnable-solution-by-ssrlive-wh5y | Code\n\nimpl Solution {\n pub fn min_score_triangulation(values: Vec<i32>) -> i32 {\n let n = values.len();\n let mut dp = vec![vec![0; n]; n]; | ssrlive | NORMAL | 2023-01-11T08:00:54.041398+00:00 | 2023-01-11T08:00:54.041440+00:00 | 33 | false | # Code\n```\nimpl Solution {\n pub fn min_score_triangulation(values: Vec<i32>) -> i32 {\n let n = values.len();\n let mut dp = vec![vec![0; n]; n];\n for j in 2..n {\n for i in (0..j - 1).rev() {\n dp[i][j] = i32::MAX;\n for k in i + 1..j {\n ... | 1 | 0 | ['Rust'] | 0 |
minimum-score-triangulation-of-polygon | EASY MCM + TABULATION C++ CODE | SHORT & BEATS 90% | easy-mcm-tabulation-c-code-short-beats-9-qj06 | \n\n- Time complexity:\n Add your time complexity here, e.g. O(n) \n O(NNN)\n\n- Space complexity:\n Add your space complexity here, e.g. O(n) \n O(N* | anmolbtw | NORMAL | 2023-01-05T12:59:22.621818+00:00 | 2023-01-05T12:59:22.621864+00:00 | 147 | false | \n\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(N*N*N)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n O(N*N)\n# Code\n```\nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& arr) {\n int N=arr.size();\n vector... | 1 | 0 | ['Array', 'Dynamic Programming', 'Recursion', 'Memoization', 'C++'] | 0 |
minimum-score-triangulation-of-polygon | EASY MCM C++ IMPLEMENTATION | SHORT AND COMMENTED CODE | easy-mcm-c-implementation-short-and-comm-hex8 | Intuition\n Describe your first thoughts on how to solve this problem. \nJust basic MCM (Matrix chain multiplication) concept implementation.\n\n\n# Complexity\ | anmolbtw | NORMAL | 2023-01-05T12:52:57.979298+00:00 | 2023-01-05T12:52:57.979343+00:00 | 90 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nJust basic MCM (Matrix chain multiplication) concept implementation.\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(N*N*N)\n\n- Space complexity:\n<!-- Add your space complexity here, ... | 1 | 0 | ['Array', 'Dynamic Programming', 'C++'] | 0 |
minimum-score-triangulation-of-polygon | Python Simple DP Solution | Faster than 94% | python-simple-dp-solution-faster-than-94-348e | Approach\n Describe your approach to solving the problem. \nWe can solve it with Dynamic programming. DP(pos1,pos2) is the minimum cost of triangulation of vert | hemantdhamija | NORMAL | 2022-12-15T09:46:34.500333+00:00 | 2022-12-15T09:46:34.500375+00:00 | 236 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nWe can solve it with Dynamic programming. ```DP(pos1,pos2)``` is the minimum cost of triangulation of vertices from pos1 to pos2. ```if (pos2-pos1<2) return 0``` means its not possible to get any triangle. Else, we do ```DP(pos1,pos2)= min(DP(pos1,pos... | 1 | 0 | ['Array', 'Dynamic Programming', 'Recursion', 'Python', 'Python3'] | 0 |
minimum-score-triangulation-of-polygon | Java || 3 approaches || Recursion, Memoization, Tabulation || Easy Understanding | java-3-approaches-recursion-memoization-xtrpv | ```\n//RECURSIVE SOLUTION\nclass Solution {\n public int minScoreTriangulation(int[] values) {\n int n = values.length;\n return recursion(valu | black_butler | NORMAL | 2022-09-13T17:21:37.083077+00:00 | 2022-09-13T17:21:37.083126+00:00 | 66 | false | ```\n//RECURSIVE SOLUTION\nclass Solution {\n public int minScoreTriangulation(int[] values) {\n int n = values.length;\n return recursion(values,0,n-1);\n }\n public int recursion(int[] v,int i,int j)\n {\n if(i+1==j) //if only two vertices are present\n return 0;\n... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Java'] | 0 |
minimum-score-triangulation-of-polygon | python, dynamic programming solution with explanation | python-dynamic-programming-solution-with-hpl2 | dp[i][j] means min score get from values[i:j+1] -> closed interval[i, j]\ndp[i][j] = (1) + (2) + (3) =min(dp[i][k] + dp[k][j] + values[i] * values[k] * values[j | shun6096tw | NORMAL | 2022-08-26T10:01:06.541128+00:00 | 2022-08-27T08:02:24.049655+00:00 | 101 | false | ```dp[i][j]``` means min score get from ```values[i:j+1]``` -> closed interval```[i, j]```\n```dp[i][j] = (1) + (2) + (3) =min(dp[i][k] + dp[k][j] + values[i] * values[k] * values[j] for k in closed interval [i+1,j-1])```\n```\n j i\n --------------\n\t / \\ | \\\n / \\ (3) | ... | 1 | 0 | ['Dynamic Programming', 'Python'] | 0 |
minimum-score-triangulation-of-polygon | C++ with Dynamic Programing | c-with-dynamic-programing-by-sroy04560-utz9 | \n### 8 ms, faster than 81.81%\n### 7.9 MB, less than 99.32% \n\nclass Solution {\npublic:\n //MCM memorization method\n // assign a matrix -1 initally\n | sroy04560 | NORMAL | 2022-08-15T20:37:44.493978+00:00 | 2022-08-15T20:37:44.494030+00:00 | 196 | false | ```\n### 8 ms, faster than 81.81%\n### 7.9 MB, less than 99.32% \n\nclass Solution {\npublic:\n //MCM memorization method\n // assign a matrix -1 initally\n //then we store value after check that t[i][j]!=-1\n int t[50][51];\n int solve(vector<int>& values,int i,int j ){\n if(i>=j)return 0;\n ... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'C'] | 0 |
minimum-score-triangulation-of-polygon | C++ | TOP DOWN DP | BOTTOM UP DP | c-top-down-dp-bottom-up-dp-by-sharmaachi-jw5p | TOP DOWN APPROACH\nTime Complexity -> O(N)\nSpace Complexity -> O (N * N)\n\n\tclass Solution {\n\tpublic:\n \n int solveMem(vector &v, int i, int j, vect | sharmaachintya | NORMAL | 2022-07-18T10:56:18.575196+00:00 | 2022-07-18T10:56:18.575243+00:00 | 26 | false | **TOP DOWN APPROACH**\n*Time Complexity ->* O(N)\n*Space Complexity ->* O (N * N)\n\n\tclass Solution {\n\tpublic:\n \n int solveMem(vector<int> &v, int i, int j, vector<vector<int>> &dp)\n {\n // base case\n if (i+1 == j)\n return 0;\n \n if(dp[i][j] != -1)\n ... | 1 | 0 | ['Dynamic Programming'] | 0 |
minimum-score-triangulation-of-polygon | Very Easy clean code | Variation of MCM | very-easy-clean-code-variation-of-mcm-by-hanc | If you have seen Balloon Burst and this problem, not able to find the solution .\nJust read the following pattern carefully .\n\nThese both are the child proble | AjayRajawat01 | NORMAL | 2022-07-15T05:52:14.380382+00:00 | 2022-07-15T05:52:14.380420+00:00 | 55 | false | If you have seen Balloon Burst and this problem, not able to find the solution .\nJust read the following pattern carefully .\n\nThese both are the child problem of MCM .\n\nIf you have seen Balloon Burst and this problem, not able to find the solution .\nJust read the following pattern carefully .\n\nThese both are th... | 1 | 0 | [] | 0 |
minimum-score-triangulation-of-polygon | JAVA|| HARD TO UNDERSTAD CODE | java-hard-to-understad-code-by-ankit1104-1sww | class Solution {\n public int minScoreTriangulation(int[] arr) {\n int n=arr.length;\n int dp[][] = new int[101][101];\n for(int i=0;i<1 | ankit1104 | NORMAL | 2022-06-29T21:07:44.847938+00:00 | 2022-06-29T21:07:44.847970+00:00 | 52 | false | class Solution {\n public int minScoreTriangulation(int[] arr) {\n int n=arr.length;\n int dp[][] = new int[101][101];\n for(int i=0;i<101;i++){\n for(int j=0;j<101;j++){\n dp[i][j]=-1;\n }\n }\n return solve(arr,1,n-1,dp);\n \n }\n ... | 1 | 0 | [] | 0 |
minimum-score-triangulation-of-polygon | C++✅ || Simple || Memoization code | c-simple-memoization-code-by-prinzeop-xw0c | \nclass Solution {\nprivate:\n\tint helper(vector<int> &nums, int i, int j, vector<vector<int>> &dp) {\n\t\tif (j - i <= 1) return 0; // no triangle\n\t\tif (dp | casperZz | NORMAL | 2022-06-17T18:14:34.978945+00:00 | 2022-06-17T18:14:34.978986+00:00 | 101 | false | ```\nclass Solution {\nprivate:\n\tint helper(vector<int> &nums, int i, int j, vector<vector<int>> &dp) {\n\t\tif (j - i <= 1) return 0; // no triangle\n\t\tif (dp[i][j] != -1) return dp[i][j];\n\t\tint mini = 1e9;\n\t\tfor (int ind = i + 1; ind < j; ++ind) {\n\t\t\tint cost = nums[i] * nums[ind] * nums[j] + helper(num... | 1 | 0 | ['C'] | 0 |
minimum-score-triangulation-of-polygon | JavaScript Recursive: 60% time, 46% space | javascript-recursive-60-time-46-space-by-5udp | \nvar minScoreTriangulation = function(values) {\n let dp = Array(values.length).fill().map((i) => Array(values.length).fill(0));\n function dfs(i, j) {\n | hqz3 | NORMAL | 2022-05-29T23:30:04.992116+00:00 | 2022-05-29T23:30:04.992148+00:00 | 126 | false | ```\nvar minScoreTriangulation = function(values) {\n let dp = Array(values.length).fill().map((i) => Array(values.length).fill(0));\n function dfs(i, j) {\n if (dp[i][j]) return dp[i][j];\n if (j - i < 2) return 0;\n let min = Infinity;\n // k forms a triangle with i and j, thus bisec... | 1 | 0 | ['JavaScript'] | 1 |
minimum-score-triangulation-of-polygon | Matrix Chain Multiplication || Easy || C++ || Memoization | matrix-chain-multiplication-easy-c-memoi-0r5v | \nclass Solution {\npublic:\n \n int dp[51][51];\n \n int solve(vector<int>&values, int i, int j)\n {\n if(i>=j)\n return 0;\n | i_m_harsh_shah | NORMAL | 2022-05-28T12:28:38.572971+00:00 | 2022-05-28T12:28:38.573018+00:00 | 130 | false | ```\nclass Solution {\npublic:\n \n int dp[51][51];\n \n int solve(vector<int>&values, int i, int j)\n {\n if(i>=j)\n return 0;\n if(dp[i][j]!=-1)\n return dp[i][j];\n \n int ans = INT_MAX;\n \n for(int k=i;k<=j-1;k++)\n {\n ... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'C', 'C++'] | 0 |
minimum-score-triangulation-of-polygon | Clean and concise MCM code || Recursion + Memorization || C++ | clean-and-concise-mcm-code-recursion-mem-692n | \nclass Solution {\nprivate:\n int solve(int st,int en,vector<int> &arr,vector<vector<int>> &dp){\n if(st+1 == en) return 0;\n if(dp[st][en] != | _Pinocchio | NORMAL | 2022-05-19T16:09:56.224890+00:00 | 2022-05-19T16:09:56.224934+00:00 | 143 | false | ```\nclass Solution {\nprivate:\n int solve(int st,int en,vector<int> &arr,vector<vector<int>> &dp){\n if(st+1 == en) return 0;\n if(dp[st][en] != -1) return dp[st][en];\n \n int ans = 1e9;\n \n for(int cut=st+1;cut<en;cut++){\n int left = solve(st,cut,arr,dp);\n ... | 1 | 0 | ['Recursion', 'Memoization', 'C', 'C++'] | 0 |
minimum-score-triangulation-of-polygon | PYTHON SOL || RECURSION + MEMO || EXPLAINED || WITH PICTURES || | python-sol-recursion-memo-explained-with-yfwz | EXPLAINED\n\nWe need to make n-2 triangles -> For this we need to remove n -3 triangles from polygon\n\nNow a polygon with side say 6 can have multiple ways to | reaper_27 | NORMAL | 2022-05-13T13:03:56.810263+00:00 | 2022-05-13T13:03:56.810294+00:00 | 172 | false | # EXPLAINED\n```\nWe need to make n-2 triangles -> For this we need to remove n -3 triangles from polygon\n\nNow a polygon with side say 6 can have multiple ways to remove triangle ( multiple triangle)\n\nSo here comes DP\nWe try each traingle ony by one\n\n```\n\n {\n int n=values.size();\n \n vector<vector<int>> ca | cx3129 | NORMAL | 2022-04-21T05:16:39.115354+00:00 | 2022-04-21T05:17:36.797004+00:00 | 117 | false | ```\n\nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& values)\n {\n int n=values.size();\n \n vector<vector<int>> cache(n,vector<int>(n,-1));\n return calculate(values, 0, n - 1, cache);\n }\n\n int calculate(vector<int>& values, int from, int to,vector<vector... | 1 | 0 | [] | 1 |
minimum-score-triangulation-of-polygon | TypeScript/JavaScript Dynamic Programming Solution | typescriptjavascript-dynamic-programming-xljo | \nconst minScoreTriangulation = (values: number[]): number => {\n const dp = new Array(values.length + 1)\n .fill(0)\n .map(() => new Array(values.length | the-technomancer | NORMAL | 2022-04-10T08:18:07.981493+00:00 | 2022-04-10T08:18:07.981531+00:00 | 110 | false | ```\nconst minScoreTriangulation = (values: number[]): number => {\n const dp = new Array(values.length + 1)\n .fill(0)\n .map(() => new Array(values.length + 1).fill(null));\n\n return minScoreTriangulationHelper(values, dp, 0, values.length - 1);\n};\n\nconst minScoreTriangulationHelper = (\n values: number[... | 1 | 0 | ['Dynamic Programming', 'TypeScript', 'JavaScript'] | 0 |
minimum-score-triangulation-of-polygon | c++ || DP || memoization | c-dp-memoization-by-jyotirmayjain_27-qrym | \nclass Solution {\npublic:\n int dp[60][60];\n int mcm(vector<int>&v,int i,int j)\n {\n if(dp[i][j]!=-1)\n return dp[i][j];\n | jyotirmayjain_27 | NORMAL | 2022-04-01T08:09:48.795188+00:00 | 2022-04-01T08:09:48.795221+00:00 | 106 | false | ```\nclass Solution {\npublic:\n int dp[60][60];\n int mcm(vector<int>&v,int i,int j)\n {\n if(dp[i][j]!=-1)\n return dp[i][j];\n int ans=INT_MAX;\n for(int k=i+1;k<j;k++)\n {\n int temp=v[i]*v[k]*v[j]+mcm(v,i,k)+mcm(v,k,j);\n // cout<<temp<<" " << ... | 1 | 0 | ['Dynamic Programming', 'Memoization'] | 0 |
minimum-score-triangulation-of-polygon | PYTHON solution Memoization DP | python-solution-memoization-dp-by-raghav-ub5y | \t#MEMOIZATION\t\n\t\tclass Solution:\n\t\t\tdef minScoreTriangulation(self, arr: List[int]) -> int:\n\t\t\t\tN=len(arr)\n\t\t\t\tdp=[[-1](N+1) for i in range(N | RaghavGupta22 | NORMAL | 2022-02-22T06:26:38.170682+00:00 | 2022-02-22T06:26:38.170725+00:00 | 245 | false | \t#MEMOIZATION\t\n\t\tclass Solution:\n\t\t\tdef minScoreTriangulation(self, arr: List[int]) -> int:\n\t\t\t\tN=len(arr)\n\t\t\t\tdp=[[-1]*(N+1) for i in range(N+1)]\n\t\t\t\tdef solve(i,j):\n\t\t\t\t\tif i>=j:\n\t\t\t\t\t\treturn 0 \n\t\t\t\t\tif dp[i][j]!=-1:\n\t\t\t\t\t\treturn dp[i][j]\n\t\t\t\t\tminn=float(\'in... | 1 | 0 | ['Dynamic Programming', 'Memoization', 'Python'] | 0 |
minimum-score-triangulation-of-polygon | Just Matrix Chain Multiplication 😋!!! | just-matrix-chain-multiplication-by-chan-uk29 | Just observe the test cases and your calculated output and do keep in mind the problem of matrix chain multiplication. You will find that it\'s basically exactl | chandramani_lc | NORMAL | 2022-02-16T06:34:22.494477+00:00 | 2022-02-19T11:54:26.533678+00:00 | 204 | false | Just observe the test cases and your calculated output and do keep in mind the problem of matrix chain multiplication. You will find that it\'s basically exactly the same.\n\n```\nclass Solution {\npublic:\n int dp[51][51];\n \n int help(vector<int>& values, int l, int r)\n {\n if(l >= r)\n ... | 1 | 0 | ['C', 'Matrix', 'C++'] | 0 |
minimum-score-triangulation-of-polygon | C++ RECURSION | MEMOIZATION | CATALAN NUMBERS | c-recursion-memoization-catalan-numbers-dt45l | Simple Recursive Solution with Memoization\n```\nclass Solution {\npublic:\n vector> dp;\n int fun(int i,int j,vector &arr){\n //Base case \n | njcoder | NORMAL | 2022-02-10T06:54:01.504487+00:00 | 2022-02-10T06:54:01.504523+00:00 | 141 | false | **Simple Recursive Solution with Memoization**\n```\nclass Solution {\npublic:\n vector<vector<int>> dp;\n int fun(int i,int j,vector<int> &arr){\n //Base case \n if(j - i <= 1) return 0;\n if(j - i == 2) return arr[i]*arr[i+1]*arr[i+2];\n if(dp[i][j] != -1) return dp[i][j];\n i... | 1 | 0 | ['Recursion', 'Memoization'] | 0 |
minimum-score-triangulation-of-polygon | [python] top down dp with comments | python-top-down-dp-with-comments-by-somb-xtnb | \n"""\n\n1039. Minimum Score Triangulation of Polygon\n\nVery similiar to other problems on DP on intervals (aka MCM). You don\'t particularily need\nthe geomet | somb | NORMAL | 2022-02-08T16:37:01.923598+00:00 | 2022-02-08T16:37:01.923640+00:00 | 139 | false | ```\n"""\n\n1039. Minimum Score Triangulation of Polygon\n\nVery similiar to other problems on DP on intervals (aka MCM). You don\'t particularily need\nthe geometric intuition for this one. \n\nI copy/pasted my soln for Burst Balloons and changed the following things:\nhttps://leetcode.com/problems/burst-balloons/disc... | 1 | 0 | ['Dynamic Programming', 'Python'] | 0 |
minimum-score-triangulation-of-polygon | clean and easy c++ solution || 4ms runtime | clean-and-easy-c-solution-4ms-runtime-by-y9xw | \nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& arr) {\n int n = arr.size();\n int dp[n-1][n-1];\n for(int g = 0; g < n-1; g++) | thanoschild | NORMAL | 2022-02-03T06:59:19.676789+00:00 | 2022-02-03T06:59:19.676827+00:00 | 95 | false | ```\nclass Solution {\npublic:\n int minScoreTriangulation(vector<int>& arr) {\n int n = arr.size();\n int dp[n-1][n-1];\n for(int g = 0; g < n-1; g++)\n {\n for(int i=0, j = g; j<n-1; i++, j++)\n {\n if(g == 0)\n dp[i][j] = 0;\n else if(g == 1)\n dp[i]... | 1 | 0 | ['Dynamic Programming', 'C'] | 0 |
minimum-score-triangulation-of-polygon | c++ simple solution , matrix chain multiplication | c-simple-solution-matrix-chain-multiplic-onjz | \nclass Solution {\npublic:\n int t[51][51];\n int solve(vector<int> & arr , int i , int j){\n if(i>=j){\n return 0;\n }\n | sparsh3435 | NORMAL | 2022-02-02T11:35:58.852625+00:00 | 2022-02-02T11:35:58.852668+00:00 | 76 | false | ```\nclass Solution {\npublic:\n int t[51][51];\n int solve(vector<int> & arr , int i , int j){\n if(i>=j){\n return 0;\n }\n if(t[i][j]!=-1){\n return t[i][j];\n }\n int mn = INT_MAX;\n for(int k = i ; k<j ; k++){\n int temp = solve(arr ,... | 1 | 0 | ['Dynamic Programming'] | 0 |
minimum-score-triangulation-of-polygon | C++ | Gap Strategy | Matrix Chain Multiplication | (Tabulation+Memoization) | c-gap-strategy-matrix-chain-multiplicati-defu | Recursion+Memoization\n\n//TC===>O(n^3)\n//SC====>O(51^2)\nint dp[51][51];\nint solve(int i,int j,vector<int>&values)\n{\n if(j-i<2)return 0; \n \n \n | utsav___gupta | NORMAL | 2022-01-31T06:01:38.615211+00:00 | 2022-01-31T06:22:01.136813+00:00 | 119 | false | **Recursion+Memoization**\n```\n//TC===>O(n^3)\n//SC====>O(51^2)\nint dp[51][51];\nint solve(int i,int j,vector<int>&values)\n{\n if(j-i<2)return 0; \n \n \n if(dp[i][j]!=-1)return dp[i][j];\n int ans=INT_MAX;\n for(int k=i+1;k<j;k++)\n {\n int v1=values[i]*values[j]*values[k];\n int... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'C'] | 0 |
minimum-score-triangulation-of-polygon | Easy Java DP Solution | easy-java-dp-solution-by-parikshit3097-crec | \nclass Solution {\n public int minScoreTriangulation(int[] arr) {\n if(arr.length ==3){\n return arr[0] * arr[1] * arr[2];\n }\n | parikshit3097 | NORMAL | 2022-01-25T02:16:21.058945+00:00 | 2022-01-25T02:16:21.058981+00:00 | 67 | false | ```\nclass Solution {\n public int minScoreTriangulation(int[] arr) {\n if(arr.length ==3){\n return arr[0] * arr[1] * arr[2];\n }\n int [][]dp = new int[arr.length][arr.length];\n \n for(int gap = 2; gap<arr.length; gap++){\n for(int left=0; left<arr.length-g... | 1 | 0 | [] | 0 |
minimum-score-triangulation-of-polygon | Same as Matrix Chain Multiplication || Standard Problem || Memoized Code | same-as-matrix-chain-multiplication-stan-9g5g | class Solution {\npublic:\n \n int t[101][101];\n \n int solve(vector& arr, int i, int j)\n {\n //memoized code\n if(i>=j)\n | Sumit4399 | NORMAL | 2021-12-31T16:30:34.777140+00:00 | 2021-12-31T16:30:34.777170+00:00 | 107 | false | class Solution {\npublic:\n \n int t[101][101];\n \n int solve(vector<int>& arr, int i, int j)\n {\n //memoized code\n if(i>=j)\n return 0;\n \n if(t[i][j] != -1)\n return t[i][j];\n \n int mn= INT_MAX;\n for(int k=i; k<j; k++)\n {\n ... | 1 | 1 | ['Dynamic Programming', 'C'] | 0 |
minimum-score-triangulation-of-polygon | Most Easy Memo in the Universe | most-easy-memo-in-the-universe-by-mayank-qjxy | I STRONGLY URGE YOU TO SEE THE TABULATION SOLUTION FIRST (IF INDIAN THEN FROM PEPCODING YOUTUBE CHANNEL)\n\n\nclass Solution {\n public int minScoreTriangula | mayank357000 | NORMAL | 2021-12-04T12:31:02.094734+00:00 | 2021-12-04T12:31:02.094761+00:00 | 242 | false | I STRONGLY URGE YOU TO SEE THE TABULATION SOLUTION FIRST (IF INDIAN THEN FROM PEPCODING YOUTUBE CHANNEL)\n\n```\nclass Solution {\n public int minScoreTriangulation(int[] arr) {\n int dp[][]=new int[arr.length][arr.length];\n for(int i=0;i<arr.length;i++)\n {\n Arrays.fill(dp[i],-1);\n }... | 1 | 0 | ['Memoization', 'Java'] | 1 |
remove-outermost-parentheses | [Java/C++/Python] Count Opened Parenthesis | javacpython-count-opened-parenthesis-by-p45ye | Intuition\nQuote from @shubhama,\nPrimitive string will have equal number of opened and closed parenthesis.\n\n## Explanation:\nopened count the number of opene | lee215 | NORMAL | 2019-04-07T04:11:05.952839+00:00 | 2019-04-07T04:11:05.952913+00:00 | 57,070 | false | ## **Intuition**\nQuote from @shubhama,\nPrimitive string will have equal number of opened and closed parenthesis.\n\n## **Explanation**:\n`opened` count the number of opened parenthesis.\nAdd every char to the result,\nunless the first left parenthesis,\nand the last right parenthesis.\n\n## **Time Complexity**:\n`O(N... | 781 | 7 | [] | 92 |
remove-outermost-parentheses | Solution | solution-by-deleted_user-9ev5 | C++ []\nclass Solution {\npublic:\n string removeOuterParentheses(string S) {\n string res;\n int opened = 0;\n for (char c : S) {\n | deleted_user | NORMAL | 2023-05-22T07:17:34.191764+00:00 | 2023-05-22T07:47:24.600942+00:00 | 42,627 | false | ```C++ []\nclass Solution {\npublic:\n string removeOuterParentheses(string S) {\n string res;\n int opened = 0;\n for (char c : S) {\n if (c == \'(\' && opened++ > 0) res += c;\n if (c == \')\' && opened-- > 1) res += c;\n }\n return res;\n }\n};\n```\n\n`... | 500 | 1 | ['C++', 'Java', 'Python3'] | 13 |
remove-outermost-parentheses | ✅ Beats 100 % || C++ || Easy to Understand || Without Using Stack | beats-100-c-easy-to-understand-without-u-z586 | Intuition\n Describe your first thoughts on how to solve this problem. \nWe need to remove the outermost parentheses from each segment of valid parentheses in t | chitrakshsuri | NORMAL | 2024-06-04T06:32:49.321582+00:00 | 2024-06-04T06:32:49.321603+00:00 | 25,463 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe need to remove the outermost parentheses from each segment of valid parentheses in the given string. A valid parentheses string is one where every opening parenthesis ( has a matching closing parenthesis ). If we can track how many ope... | 380 | 1 | ['C++'] | 6 |
remove-outermost-parentheses | Well Explained Code in JAVA || | well-explained-code-in-java-by-sakshamka-45oe | \n\n# Approach\nThis is a solution to the problem of removing outermost parentheses from a string containing only parentheses.\n\nThe approach used is to keep t | sakshamkaushiik | NORMAL | 2023-02-09T19:11:43.729664+00:00 | 2023-02-09T19:11:43.729710+00:00 | 17,848 | false | \n\n# Approach\nThis is a solution to the problem of removing outermost parentheses from a string containing only parentheses.\n\nThe approach used is to keep track of the parentheses using a stack. Whenever an opening parenthesis is encountered, it is pushed onto the stack. Whenever a closing parenthesis is encountere... | 246 | 0 | ['Stack', 'Java'] | 6 |
remove-outermost-parentheses | My Java 3ms Straight Forward Solution | Beats 100% | my-java-3ms-straight-forward-solution-be-fnze | \nclass Solution {\n public String removeOuterParentheses(String S) {\n \n StringBuilder sb = new StringBuilder();\n int open=0, close=0 | debdattakunda | NORMAL | 2019-04-07T19:53:33.359919+00:00 | 2019-04-07T19:53:33.359985+00:00 | 13,058 | false | ```\nclass Solution {\n public String removeOuterParentheses(String S) {\n \n StringBuilder sb = new StringBuilder();\n int open=0, close=0, start=0;\n for(int i=0; i<S.length(); i++) {\n if(S.charAt(i) == \'(\') {\n open++;\n } else if(S.charAt(i) == ... | 109 | 3 | [] | 15 |
remove-outermost-parentheses | C++ 0ms solution | c-0ms-solution-by-dan0907-62w1 | \nclass Solution {\npublic:\n string removeOuterParentheses(string S) {\n int count = 0;\n std::string str;\n for (char c : S) {\n | dan0907 | NORMAL | 2019-05-10T17:29:30.113941+00:00 | 2019-10-29T13:14:01.405451+00:00 | 10,291 | false | ```\nclass Solution {\npublic:\n string removeOuterParentheses(string S) {\n int count = 0;\n std::string str;\n for (char c : S) {\n if (c == \'(\') {\n if (count++) {\n str += \'(\';\n }\n } else {\n if (--co... | 98 | 1 | ['C'] | 11 |
remove-outermost-parentheses | 💡One Pass | FAANG SDE-1 Interview😯 | one-pass-faang-sde-1-interview-by-aditya-m4il | \nVery Easy Beginner Friendly Solution \uD83D\uDCA1\nDo not use stack to prevent more extra space.\n\n\nPlease do Upvote if it helps :)\n\n\n# Complexity\n- Tim | AdityaBhate | NORMAL | 2022-12-16T06:40:07.118768+00:00 | 2022-12-16T06:40:07.118804+00:00 | 16,202 | false | ```\nVery Easy Beginner Friendly Solution \uD83D\uDCA1\nDo not use stack to prevent more extra space.\n\n\nPlease do Upvote if it helps :)\n```\n\n# Complexity\n- Time complexity: O(n) //One pass\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n) //For resultant string\n<!-- Add your spa... | 85 | 3 | ['String', 'Stack', 'C++', 'Java', 'Python3'] | 7 |
remove-outermost-parentheses | [c++] two solution stack and with out stack(only slight modification in stack sol.) | c-two-solution-stack-and-with-out-stacko-qxfn | please upVote my solution if you like it.\n\nMy first solution is stack based and it consume more memory than with out stack solution in my second solution to e | sanjeev1709912 | NORMAL | 2020-08-28T04:47:50.579025+00:00 | 2020-08-28T04:48:12.549655+00:00 | 4,982 | false | please **upVote** my solution if you like it.\n\nMy first solution is stack based and it consume more memory than with out stack solution in my second solution to elemenate stack from it so please reffer to that also \n\n\n\'\'\'\nclass Solution {\npublic:\n\n string removeOuterParentheses(string S) {\n stack... | 63 | 0 | ['Stack', 'C'] | 3 |
remove-outermost-parentheses | [Python] Simple O(n) solution - beats 97% | python-simple-on-solution-beats-97-by-ye-5se3 | python\ndef removeOuterParentheses(self, S):\n\tres = []\n\tbalance = 0\n\ti = 0\n\tfor j in range(len(S)):\n\t\tif S[j] == "(":\n\t\t\tbalance += 1\n\t\telif S | yerbola | NORMAL | 2019-05-26T04:59:38.122390+00:00 | 2019-11-02T20:00:13.945787+00:00 | 5,265 | false | ```python\ndef removeOuterParentheses(self, S):\n\tres = []\n\tbalance = 0\n\ti = 0\n\tfor j in range(len(S)):\n\t\tif S[j] == "(":\n\t\t\tbalance += 1\n\t\telif S[j] == ")":\n\t\t\tbalance -= 1\n\t\tif balance == 0:\n\t\t\tres.append(S[i+1:j])\n\t\t\ti = j+1\n\treturn "".join(res)\n``` | 52 | 0 | [] | 10 |
remove-outermost-parentheses | Easy to understand Python with comments | easy-to-understand-python-with-comments-6qskb | Python\nclass Solution:\n def removeOuterParentheses(self, S: str) -> str:\n res = \'\'\n stack = []\n \n # basket is used to sto | cglotr | NORMAL | 2019-06-25T14:34:23.879552+00:00 | 2019-06-29T15:29:58.771041+00:00 | 3,814 | false | ```Python\nclass Solution:\n def removeOuterParentheses(self, S: str) -> str:\n res = \'\'\n stack = []\n \n # basket is used to store previous value\n basket = \'\'\n \n for p in S:\n if p == \'(\':\n stack.append(p)\n else:\n ... | 47 | 0 | ['Stack', 'Python'] | 4 |
remove-outermost-parentheses | [JAVA] beats 98%, simple iterative solution | java-beats-98-simple-iterative-solution-n33eq | \nclass Solution {\n public String removeOuterParentheses(String S) {\n StringBuilder sb = new StringBuilder();\n int counter = 0;\n for | jamsrandorj | NORMAL | 2019-11-13T21:58:11.645114+00:00 | 2019-11-13T21:58:11.645149+00:00 | 3,143 | false | ```\nclass Solution {\n public String removeOuterParentheses(String S) {\n StringBuilder sb = new StringBuilder();\n int counter = 0;\n for(char c : S.toCharArray()){\n if(c == \'(\'){\n if(counter != 0) sb.append(c);\n counter++;\n }\n ... | 35 | 1 | ['Java'] | 4 |
remove-outermost-parentheses | Ridiculously Simple JAVA O(n) Solution + Explanation [0ms Beats 100% Time & Memory] | ridiculously-simple-java-on-solution-exp-tmav | Explanation:\nSince the input String only consists of parentheses we don\'t even have to mainatain a Stack. We can simply maintain a counter which is O(1) and k | agrawroh | NORMAL | 2019-04-12T19:20:05.924040+00:00 | 2019-04-12T19:20:05.924136+00:00 | 4,390 | false | **Explanation:**\nSince the input String only consists of parentheses we don\'t even have to mainatain a Stack. We can simply maintain a counter which is O(1) and keep incrementing and decrementing it\'s value based on the opening/closing bracket.\n\n**Algorithm:**\n1. Convert the given input String to a `char` array a... | 32 | 0 | [] | 4 |
remove-outermost-parentheses | C++ Two pointers | c-two-pointers-by-votrubac-hq23 | Intuition\nWhen the number of open parentheses equals closed, we found a primitive string.\n# Solution\nUse two pointers to track primitive strings; when open = | votrubac | NORMAL | 2019-04-07T04:01:04.880510+00:00 | 2019-04-07T04:01:04.880544+00:00 | 4,220 | false | # Intuition\nWhen the number of ```open``` parentheses equals ```closed```, we found a primitive string.\n# Solution\nUse two pointers to track primitive strings; when ```open == close```, remove outermost parentheses and add the string to the result.\n```\nstring removeOuterParentheses(string S, string res = "") {\n ... | 31 | 3 | [] | 4 |
remove-outermost-parentheses | Python - Super Easy - 98% Speed | python-super-easy-98-speed-by-aragorn-8wku | We just need a For-Loop to count the number of Parenthesis open. The "append" operator goes at the center of the expression to avoid including the Outermost Pat | aragorn_ | NORMAL | 2020-04-24T00:48:34.579703+00:00 | 2020-07-01T02:31:33.295675+00:00 | 3,796 | false | We just need a For-Loop to count the number of Parenthesis open. The "append" operator goes at the center of the expression to avoid including the Outermost Patentheses. Cheers,\n\n```\nclass Solution:\n def removeOuterParentheses(self, S: str) -> str:\n popen, result = 0, []\n for x in S:\n ... | 30 | 0 | ['Python', 'Python3'] | 2 |
remove-outermost-parentheses | Python solution using stack and maintaining counter | python-solution-using-stack-and-maintain-8bcy | Before I explain any further get this, let\'s say for each "(" you get you put -1 and for each ")" you get you put +1 to the counter varriable and because of th | dubeyaman157 | NORMAL | 2022-10-09T07:50:10.472671+00:00 | 2022-12-25T08:26:21.686114+00:00 | 1,725 | false | # Before I explain any further get this, let\'s say for each "(" you get you put -1 and for each ")" you get you put +1 to the counter varriable and because of that whenever we encounter a valid paranthese our sum will be zero for example for (()())(()) can be decomposed to (()()) + (()) note that for each valid decomp... | 25 | 0 | ['Stack', 'Python'] | 2 |
remove-outermost-parentheses | ✅Java Simple Solution || ✅Runtime 2ms || ✅ Beats100% | java-simple-solution-runtime-2ms-beats10-m5lq | 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 | ahmedna126 | NORMAL | 2023-08-29T19:41:17.745408+00:00 | 2023-11-07T11:40:55.689863+00:00 | 2,206 | false | # 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. $$O(n)$$ --... | 23 | 0 | ['Java'] | 2 |
remove-outermost-parentheses | C++ stack and without stack solutions. | c-stack-and-without-stack-solutions-by-k-hvf6 | Stack implementation\n\nclass Solution {\npublic:\n string removeOuterParentheses(string s)\n {\n stack<char>sc;\n string ans;\n for( | kfaisal-se | NORMAL | 2021-06-10T10:35:12.857655+00:00 | 2021-06-10T10:35:12.857687+00:00 | 2,968 | false | **Stack implementation**\n```\nclass Solution {\npublic:\n string removeOuterParentheses(string s)\n {\n stack<char>sc;\n string ans;\n for(char i:s)\n {\n if(i == \'(\')\n {\n if(sc.size() > 0)\n {\n ans += i;\n ... | 23 | 0 | ['Stack', 'C', 'C++'] | 2 |
remove-outermost-parentheses | Javascript solution - 98% faster | javascript-solution-98-faster-by-whiskey-t3pm | \nvar removeOuterParentheses = function(S) {\n let parenthesCount = 0;\n let result = "";\n \n for (const letter of S) {\n if (letter === "(" | whiskey022 | NORMAL | 2019-09-11T08:57:39.284718+00:00 | 2019-09-11T09:01:16.318939+00:00 | 2,705 | false | ```\nvar removeOuterParentheses = function(S) {\n let parenthesCount = 0;\n let result = "";\n \n for (const letter of S) {\n if (letter === "(") {\n if (parenthesCount) {\n result += letter;\n }\n parenthesCount++;\n } else {\n parent... | 22 | 0 | ['JavaScript'] | 2 |
remove-outermost-parentheses | Beats 100% 🔥 of users|| JAVA || Without Using Stack || Easy to understand ✅ | beats-100-of-users-java-without-using-st-0n68 | Intuition\n Describe your first thoughts on how to solve this problem. \nRemove the outermost parentheses from each primitive valid parentheses substring by tra | Abhishek_Yadav_leetcode | NORMAL | 2024-09-07T19:56:23.427996+00:00 | 2024-09-08T08:08:02.534349+00:00 | 1,772 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nRemove the outermost parentheses from each primitive valid parentheses substring by tracking balance.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1) Initialize: Convert the string to a character array, create a... | 19 | 0 | ['Java'] | 2 |
remove-outermost-parentheses | Javascript beats 99.26% easy to understand | javascript-beats-9926-easy-to-understand-abi7 | \nvar removeOuterParentheses = function(S) {\n let result = \'\';\n let open = 0\n for (let i = 0; i < S.length; i++) {\n if (S[i] === \'(\') {\ | careyl96 | NORMAL | 2019-05-29T00:01:24.199522+00:00 | 2019-05-29T00:01:24.199590+00:00 | 1,048 | false | ```\nvar removeOuterParentheses = function(S) {\n let result = \'\';\n let open = 0\n for (let i = 0; i < S.length; i++) {\n if (S[i] === \'(\') {\n if (open > 0) { \n\t\t\t\tresult += \'(\';\n\t\t\t}\n\t\t\topen++;\n } else if (S[i] === \')\') {\n if (open > 1) { \n\t\t\t\t... | 18 | 0 | [] | 0 |
remove-outermost-parentheses | Python3- simple solution 99.8% | python3-simple-solution-998-by-logan_kd-cb6p | \nclass Solution:\n def removeOuterParentheses(self, S: str) -> str:\n res = ""\n count = 0\n first = 0\n for i in range(len(S)): | logan_kd | NORMAL | 2019-10-06T22:58:38.602821+00:00 | 2019-10-06T23:05:49.107911+00:00 | 2,312 | false | ```\nclass Solution:\n def removeOuterParentheses(self, S: str) -> str:\n res = ""\n count = 0\n first = 0\n for i in range(len(S)):\n if S[i] == "(":\n count +=1\n else:\n count -= 1\n \n if(count == 0):\n ... | 17 | 0 | ['Python', 'Python3'] | 5 |
remove-outermost-parentheses | Shortest Python Solution | shortest-python-solution-by-flowingwater-l1cx | \nclass Solution:\n def removeOuterParentheses(self, S: str) -> str:\n cnt, res = 0, \'\'\n for c in S:\n if c == \')\': cnt -= 1 \ | flowingwater526 | NORMAL | 2019-08-14T06:47:41.356265+00:00 | 2019-08-14T06:47:41.356330+00:00 | 2,852 | false | ```\nclass Solution:\n def removeOuterParentheses(self, S: str) -> str:\n cnt, res = 0, \'\'\n for c in S:\n if c == \')\': cnt -= 1 \n if cnt != 0: res += c \n if c == \'(\': cnt+=1 \n return res\n``` \n | 16 | 0 | ['Python', 'Python3'] | 5 |
remove-outermost-parentheses | ✅☑️ Best C++ 2 Solution Ever || String || Stack || One Stop Solution. | best-c-2-solution-ever-string-stack-one-rq3hx | Intuition\n Describe your first thoughts on how to solve this problem. \nWe can solve this question using Multiple Approaches. (Here I have explained all the po | its_vishal_7575 | NORMAL | 2023-02-16T17:03:13.887148+00:00 | 2023-02-16T17:03:13.887182+00:00 | 2,410 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can solve this question using Multiple Approaches. (Here I have explained all the possible solutions of this problem).\n\n1. Solved using String + Stack.\n2. Solved using String.\n\n# Approach\n<!-- Describe your approach to solving th... | 15 | 0 | ['String', 'Stack', 'C++'] | 1 |
remove-outermost-parentheses | Easy Solution with Dry Run and Example | easy-solution-with-dry-run-and-example-b-xotj | Intuition\n Describe your first thoughts on how to solve this problem. \nTo solve this, we need to:\n\n1. Traverse the string while keeping track of the number | reaperrrrrr | NORMAL | 2024-08-24T17:02:28.139784+00:00 | 2024-08-24T17:02:28.139809+00:00 | 1,128 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve this, we need to:\n\n1. Traverse the string while keeping track of the number of open and close parentheses using a counter.\n \n1. Append the parentheses to the result string only when they are not the outermost ones.\n\n---\n\n... | 13 | 0 | ['String', 'Python', 'C++', 'Java', 'JavaScript'] | 0 |
remove-outermost-parentheses | aam admi approach | aam-admi-approach-by-tejaswibhagat-xlfg | 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 | tejaswibhagat | NORMAL | 2024-04-19T16:09:08.161290+00:00 | 2024-04-19T16:09:08.161315+00:00 | 818 | false | # 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. $$O(n)$$ --... | 13 | 0 | ['C++'] | 2 |
remove-outermost-parentheses | wont get a more straight forward solution than this. | wont-get-a-more-straight-forward-solutio-vlf4 | Intuition\n Describe your first thoughts on how to solve this problem. \nguys please upvote , I work so hard explaining things and you potatoes dont even upvote | Abhishekkant135 | NORMAL | 2024-04-14T19:31:00.969475+00:00 | 2024-04-14T19:31:00.969501+00:00 | 875 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nguys please upvote , I work so hard explaining things and you potatoes dont even upvote the post . SHAME on you. NOW UPVOTE\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n\n1. **Initializing Variables:**\n - `St... | 12 | 0 | ['String', 'Java'] | 1 |
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