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painting-a-grid-with-three-different-colors | Rust | Bottom-up DP, 6ms, beats 100% | rust-bottom-up-dp-6ms-beats-100-by-soyfl-lvza | Intuition\nCompress each row into vertices, construct a graph and do DP on it.\n\n# Complexity\n- Time complexity:\nO(n)\n\n# Code\nrust\n#[derive(Debug, Clone, | soyflourbread | NORMAL | 2024-03-05T02:13:26.747999+00:00 | 2024-03-05T02:13:26.748032+00:00 | 5 | false | # Intuition\nCompress each row into vertices, construct a graph and do DP on it.\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n# Code\n```rust\n#[derive(Debug, Clone, PartialOrd, Ord, PartialEq, Eq)]\npub enum Color {\n Blue,\n Pink,\n White,\n}\n\npub fn to_vertices(\n width: usize,\n) -> Vec<Vec<Color>... | 0 | 0 | ['Dynamic Programming', 'Rust'] | 0 |
painting-a-grid-with-three-different-colors | C++ bitmask DP | c-bitmask-dp-by-sanzenin_aria-aa0p | 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 | sanzenin_aria | NORMAL | 2024-02-21T02:40:28.726197+00:00 | 2024-02-21T02:41:34.456216+00:00 | 23 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
painting-a-grid-with-three-different-colors | DP on Counter() base (without recursion) (60% | 81%) | dp-on-counter-base-without-recursion-60-et98b | Code\n\nclass Solution:\n \n def colorTheGrid(self, m: int, n: int) -> int:\n mod = 10 ** 9 + 7\n\n def get_first_line(m) -> list[str]:\n | eshved | NORMAL | 2024-01-03T18:49:10.034401+00:00 | 2024-01-04T21:47:25.000592+00:00 | 16 | false | # Code\n```\nclass Solution:\n \n def colorTheGrid(self, m: int, n: int) -> int:\n mod = 10 ** 9 + 7\n\n def get_first_line(m) -> list[str]:\n """Returns all possible (24) variations of the first line.\n """\n res = [[], [""]]\n for i in range(m):\n ... | 0 | 0 | ['Python3'] | 0 |
painting-a-grid-with-three-different-colors | easy with all steps | easy-with-all-steps-by-parwez0786-5tjn | 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 | parwez0786 | NORMAL | 2024-01-02T12:49:33.462336+00:00 | 2024-01-02T12:49:33.462370+00:00 | 28 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
painting-a-grid-with-three-different-colors | python DP top down | python-dp-top-down-by-harrychen1995-04nc | 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 | 2023-11-13T19:25:12.504222+00:00 | 2023-11-13T19:25:12.504246+00:00 | 26 | 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. $$O(n)$$ --... | 0 | 0 | ['Dynamic Programming', 'Python3'] | 0 |
painting-a-grid-with-three-different-colors | Dynamic programming | Top to bottom | Easy to understand code | dynamic-programming-top-to-bottom-easy-t-h843 | 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 | geeknkta | NORMAL | 2023-11-01T15:13:38.051276+00:00 | 2023-11-01T15:13:38.051300+00:00 | 42 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
painting-a-grid-with-three-different-colors | C# Beats 100% in speed and memory usage | c-beats-100-in-speed-and-memory-usage-by-ug9o | \n\n\n# Complexity\n- Time complexity:\nO(n * (3^m)^2)\n\n- Space complexity:\nO(n * 3^m + (3^m)^2)\n\n# Code\n\npublic class Solution {\n public int ColorTh | junkmann | NORMAL | 2023-10-19T01:00:19.324638+00:00 | 2023-10-19T01:00:19.324672+00:00 | 7 | false | \n\n\n# Complexity\n- Time complexity:\n$$O(n * (3^m)^2)$$\n\n- Space complexity:\n$$O(n * 3^m + (3^m)^2)$$\n\n# Code\n```\npublic class Solution {\n public int ColorTheGrid(int m, int n) {\n var ... | 0 | 0 | ['C#'] | 0 |
painting-a-grid-with-three-different-colors | Java solution | java-solution-by-pejmantheory-735x | 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 | pejmantheory | NORMAL | 2023-09-16T07:48:46.672162+00:00 | 2023-09-16T07:48:46.672194+00:00 | 50 | 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. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
painting-a-grid-with-three-different-colors | Top Down dp with bit masking Easy Complete Solution | top-down-dp-with-bit-masking-easy-comple-8aqo | My first thought was to just calculate No of possible paths to n-1, m-1 with given constrain but the thing went wrong was the other left remaining squires can f | demon_code | NORMAL | 2023-09-06T08:28:03.361289+00:00 | 2023-09-06T13:02:09.588904+00:00 | 7 | false | My first thought was to just calculate No of possible paths to n-1, m-1 with given constrain but the thing went wrong was the other left remaining squires can form different combination which is not part of selected path so it failed ``` :(```\n\nto colur any cell we need information about just upper sell and left cel... | 0 | 0 | ['Dynamic Programming', 'Recursion', 'C', 'Bitmask'] | 0 |
painting-a-grid-with-three-different-colors | My Solution | my-solution-by-hope_ma-tdeg | \nclass Solution {\n public:\n int colorTheGrid(const int m, const int n) {\n constexpr int mod = 1000000007;\n constexpr int range = 2;\n constexpr i | hope_ma | NORMAL | 2023-07-12T15:01:22.873630+00:00 | 2023-07-12T15:01:22.873653+00:00 | 7 | false | ```\nclass Solution {\n public:\n int colorTheGrid(const int m, const int n) {\n constexpr int mod = 1000000007;\n constexpr int range = 2;\n constexpr int color_mask = 0b11;\n /**\n * `0b00` stands for red\n * `0b01` stands for green\n * `0b10` stands for blue\n * `0b11` is invalid\n *... | 0 | 0 | [] | 0 |
painting-a-grid-with-three-different-colors | No bitmask | no-bitmask-by-shivral-r0v2 | \n Add your space complexity here, e.g. O(n) \n\n# Code\n\nb=[]\ndef gen(s,n):\n global b\n if len(s)==n:\n fl=False\n for i in range(1,n):\ | shivral | NORMAL | 2023-05-18T13:08:26.440412+00:00 | 2023-05-18T13:08:26.440441+00:00 | 50 | false | \n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nb=[]\ndef gen(s,n):\n global b\n if len(s)==n:\n fl=False\n for i in range(1,n):\n fl|=s[i]==s[i-1]\n if not fl:\n b.append(s)\n return\n gen(s+"0",n)\n gen(s+"1",n)\n gen(s+"2",n)\n... | 0 | 0 | ['Python3'] | 0 |
painting-a-grid-with-three-different-colors | A concise solution with matrix ops only (beat 100% java submissions) | a-concise-solution-with-matrix-ops-only-i75mb | java\nclass Solution {\n public int colorTheGrid(int m, int n) {\n long mod = 1_000_000_007;\n long[][][] matrixs = {\n { { 2 } | safiir | NORMAL | 2023-05-06T07:36:44.273201+00:00 | 2023-05-06T10:07:50.976974+00:00 | 101 | false | ```java\nclass Solution {\n public int colorTheGrid(int m, int n) {\n long mod = 1_000_000_007;\n long[][][] matrixs = {\n { { 2 } },\n { { 3 } },\n { { 3, 2 }, { 2, 2 } },\n {\n { 3, 2, 1, 2 },\n ... | 0 | 0 | ['Dynamic Programming', 'Java'] | 0 |
painting-a-grid-with-three-different-colors | Python Solution | python-solution-by-aurimas13-cd3y | \n\n# Approach\nHere\'s a step-by-step explanation of the solution:\n\n1. Generate all valid rows:\nThe generate_valid_rows function generates all valid rows of | aurimas13 | NORMAL | 2023-03-15T07:20:15.852852+00:00 | 2023-03-15T07:20:15.852906+00:00 | 82 | false | \n\n# Approach\nHere\'s a step-by-step explanation of the solution:\n\n1. Generate all valid rows:\nThe generate_valid_rows function generates all valid rows of length m, with no... | 0 | 0 | ['Dynamic Programming', 'Graph', 'Python3'] | 0 |
painting-a-grid-with-three-different-colors | [Python 3] Bitmask DP | python-3-bitmask-dp-by-gabhay-2ich | Intuition\nsince m is very small so it seems sensible to use Bitmask DP here\n\n# Approach\nwe use first 5 bits for red, next 5 for blue and next 5 bits for gre | gabhay | NORMAL | 2023-02-17T06:43:04.302906+00:00 | 2023-02-17T06:48:08.630739+00:00 | 56 | false | # Intuition\nsince m is very small so it seems sensible to use Bitmask DP here\n\n# Approach\nwe use first 5 bits for red, next 5 for blue and next 5 bits for green\nand continously check for each position if we can use a specific color.\n\n# Time Compexity:\nO(n\\*m\\*2^(2\\*m))\n\n# Code\n```\nclass Solution:\n de... | 0 | 0 | ['Python3'] | 0 |
painting-a-grid-with-three-different-colors | Python recursive DP, abstracted as a Graph counting problem | python-recursive-dp-abstracted-as-a-grap-b9pk | # Intuition \n\n\n\n\n\n# Complexity\nLet w := 3^m,\n- Time complexity: O(w^2m+w^2n)\n\n\n- Space complexity: O(w^2+wn)\n\n\n# Code\n\nM = 1_000_000_007\n\n\n | vWAj0nMjOtK33vIH0jpLww | NORMAL | 2023-02-11T05:25:49.750950+00:00 | 2023-02-12T13:49:11.029090+00:00 | 57 | 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\nLet $$w := 3^m$$,\n- Time complexity: $$O(w^2m+w^2n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: ... | 0 | 0 | ['Dynamic Programming', 'Graph', 'Python3'] | 1 |
painting-a-grid-with-three-different-colors | Easy solution in c++ | easy-solution-in-c-by-doppelgangerofmeow-nxhv | Intuition\n Describe your first thoughts on how to solve this problem. \n1. Generateing all possible permutation using (m^3)\n2. Finding all permutation that co | doppelgangerOfMeow | NORMAL | 2022-12-22T11:00:50.001202+00:00 | 2022-12-22T11:00:50.001247+00:00 | 226 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. Generateing all possible permutation using (m^3)\n2. Finding all permutation that could be promoted to\na. For example, 12 -> 21, 23 and etc.\n3. Using DP to record the value of transferring\n\n# Approach\n<!-- Describe your approach t... | 0 | 0 | ['Dynamic Programming', 'Combinatorics', 'C++'] | 0 |
painting-a-grid-with-three-different-colors | Bottom-Up DP solution, beats 98% of cpp solutions | bottom-up-dp-solution-beats-98-of-cpp-so-s6e8 | Beats 98.05% in Time and 99.35% in Space for cpp solutions\n\n\n\nclass Solution {\npublic:\n const int MOD = 1e9 + 7;\n void generateRows(vector<int> &ro | mosta7il | NORMAL | 2022-07-20T19:08:54.224660+00:00 | 2022-07-20T19:08:54.224703+00:00 | 165 | false | Beats **98.05%** in Time and **99.35%** in Space for cpp solutions\n\n\n```\nclass Solution {\npublic:\n const int MOD = 1e9 + 7;\n void generateRows(vector<int> &rows, int idx, int row, int last, int &M){\... | 0 | 0 | ['Dynamic Programming'] | 0 |
painting-a-grid-with-three-different-colors | C++ Recursion + Memo | c-recursion-memo-by-dhairya_sarin-i17a | \nclass Solution {\npublic:\n string color_scheme = "RYG";\n int mod;\n vector<string>valid_moves;\n unordered_map<string,int>dp;\n void generate | dhairya_sarin | NORMAL | 2022-07-17T03:18:22.445928+00:00 | 2022-07-17T03:18:22.445956+00:00 | 180 | false | ```\nclass Solution {\npublic:\n string color_scheme = "RYG";\n int mod;\n vector<string>valid_moves;\n unordered_map<string,int>dp;\n void generateValidMoves(int n , int taken , char last, string &cur)\n { if(taken==n)\n {valid_moves.push_back(cur);\n return;\n }\n for(... | 0 | 0 | [] | 0 |
painting-a-grid-with-three-different-colors | Simple DP Implementation | simple-dp-implementation-by-day_tripper-45ea | \nclass Solution(object):\n def colorTheGrid(self, m, n):\n """\n :type m: int\n :type n: int\n :rtype: int\n """\n | Day_Tripper | NORMAL | 2022-07-03T04:03:06.582398+00:00 | 2022-07-03T04:03:06.582462+00:00 | 129 | false | ```\nclass Solution(object):\n def colorTheGrid(self, m, n):\n """\n :type m: int\n :type n: int\n :rtype: int\n """\n # m = 2\n\n patterns = ["a", "b", "c"]\n for i in range(m-1):\n new_patterns = []\n for p in patterns:\n ... | 0 | 0 | ['Dynamic Programming'] | 0 |
painting-a-grid-with-three-different-colors | This is meaningless | this-is-meaningless-by-stevefan1999-03wa | This question is just asking for the 3-coloring chromatic polynomial of the (n, m)-grid graph where n <= 5 and m <= 1000...Which is unfortunately an open proble | stevefan1999 | NORMAL | 2022-06-30T08:22:58.900665+00:00 | 2022-06-30T08:23:40.446255+00:00 | 120 | false | This question is just asking for the 3-coloring chromatic polynomial of the (n, m)-grid graph where n <= 5 and m <= 1000...Which is unfortunately an open problem.\n\nAnyone with Mathematica can brute force the formula. When the fuck did LC became an amalgamation of CodeForces and Project Euler? | 0 | 0 | [] | 1 |
painting-a-grid-with-three-different-colors | How can I improve my code. I've used backtracking but I run into TLE in some test cases | how-can-i-improve-my-code-ive-used-backt-m20w | class Solution:\n def colorTheGrid(self, m: int, n: int) -> int:\n matrix = [[0]*n for _ in range(m)]\n neighbors = [(-1,0),(0,-1)]\n de | teapea | NORMAL | 2022-06-30T04:28:57.875477+00:00 | 2022-06-30T04:28:57.875517+00:00 | 40 | false | ```class Solution:\n def colorTheGrid(self, m: int, n: int) -> int:\n matrix = [[0]*n for _ in range(m)]\n neighbors = [(-1,0),(0,-1)]\n def get_colors(r,c):\n colors = [1,2,3]\n ans = []\n if r==0 and c==0:\n return colors\n top = left ... | 0 | 0 | ['Backtracking'] | 0 |
painting-a-grid-with-three-different-colors | Python. Faster than 83%... | python-faster-than-83-by-nag007-1jpw | \nclass Solution:\n def colorTheGrid(self, m: int, n: int) -> int:\n def f(i,p=-1,s=\'\'):\n if i==m:\n res.append(s)\n | nag007 | NORMAL | 2022-06-25T16:03:23.577373+00:00 | 2022-06-25T16:03:23.577408+00:00 | 84 | false | ```\nclass Solution:\n def colorTheGrid(self, m: int, n: int) -> int:\n def f(i,p=-1,s=\'\'):\n if i==m:\n res.append(s)\n return\n for j in range(3):\n if j!=p:\n f(i+1,j,s+str(j))\n @lru_cache(None)\n def fun... | 0 | 0 | [] | 0 |
painting-a-grid-with-three-different-colors | Scala | scala-by-fairgrieve-4f02 | \nimport scala.util.chaining.scalaUtilChainingOps\n\nobject Solution {\n private object Color extends Enumeration {\n type Color = Value\n val Red, Blue, | fairgrieve | NORMAL | 2022-03-29T20:44:42.193528+00:00 | 2022-03-29T20:44:42.193554+00:00 | 53 | false | ```\nimport scala.util.chaining.scalaUtilChainingOps\n\nobject Solution {\n private object Color extends Enumeration {\n type Color = Value\n val Red, Blue, Green = Value\n }\n import Color._\n\n def colorTheGrid(m: Int, n: Int): Int = {\n val validCols = (1 to m).foldLeft(Iterable(List[Color]())) {\n ... | 0 | 0 | ['Dynamic Programming', 'Scala'] | 0 |
painting-a-grid-with-three-different-colors | Fastest O(n) Python Solution | fastest-on-python-solution-by-vkarakchee-5e0i | The solution is completely similar to recurrent one of problem 1411. Memory complexity is O(1).\n\nclass Solution:\n def colorTheGrid(self, m: int, n: int) - | vkarakcheev | NORMAL | 2022-03-11T23:44:34.452308+00:00 | 2022-03-12T10:28:05.871929+00:00 | 181 | false | The solution is completely similar to recurrent one of problem 1411. Memory complexity is O(1).\n```\nclass Solution:\n def colorTheGrid(self, m: int, n: int) -> int:\n\n if m == 1:\n return 3 * 2**(n-1) % 1000000007\n\n if m == 2:\n return 6 * 3**(n-1) % 1000000007\n\n if ... | 0 | 0 | [] | 0 |
painting-a-grid-with-three-different-colors | [Python3] DP Rolling Array & Memo | 70% Time & 95% Space | python3-dp-rolling-array-memo-70-time-95-cvt3 | \nclass Solution:\n def colorTheGrid(self, m: int, n: int) -> int:\n MOD = 10 ** 9 + 7\n # Get valid states for a column\n valid_states | pubghh01 | NORMAL | 2022-03-05T09:24:32.640384+00:00 | 2022-03-05T09:26:50.468718+00:00 | 162 | false | ```\nclass Solution:\n def colorTheGrid(self, m: int, n: int) -> int:\n MOD = 10 ** 9 + 7\n # Get valid states for a column\n valid_states = generate_states(m)\n # Init rolling array dp\n dp = [[0] * len(valid_states) for _ in range(2)]\n # Init the first column of dp to all... | 0 | 0 | ['Dynamic Programming', 'Memoization'] | 0 |
painting-a-grid-with-three-different-colors | [Rust] DP & Mask runs in 88ms | rust-dp-mask-runs-in-88ms-by-bovinovain-flex | Realizing m <= 5 is crucial. It is possible to enuemrate all valid color permutation in a column and build dp based on it.\n\nuse std::collections::HashMap;\n\n | bovinovain | NORMAL | 2022-03-01T13:50:58.215642+00:00 | 2022-03-01T13:50:58.215683+00:00 | 107 | false | Realizing m <= 5 is crucial. It is possible to enuemrate all valid color permutation in a column and build dp based on it.\n```\nuse std::collections::HashMap;\n\nconst MOD: i32 = 1000_000_007;\nimpl Solution {\n pub fn color_the_grid(m: i32, n: i32) -> i32 {\n let mask = |mut x: usize| -> Vec<usize> {\n ... | 0 | 0 | [] | 0 |
painting-a-grid-with-three-different-colors | Solution using unordered map C++ | solution-using-unordered-map-c-by-aryan2-euvs | \nclass Solution {\npublic:\n //\n int mod=1e9+7;\n char arr[3]={\'R\',\'B\',\'G\'};\n vector<string>v;\n unordered_map<string,int>ans;\n void | aryan29 | NORMAL | 2022-02-22T18:07:23.834695+00:00 | 2022-02-22T18:07:23.834737+00:00 | 190 | false | ```\nclass Solution {\npublic:\n //\n int mod=1e9+7;\n char arr[3]={\'R\',\'B\',\'G\'};\n vector<string>v;\n unordered_map<string,int>ans;\n void get_v(int m, string s)\n {\n if(m==0)\n {\n v.push_back(s);\n return;\n }\n char ch=\'$\';\n if(... | 0 | 0 | [] | 0 |
painting-a-grid-with-three-different-colors | DP with bitmask, similar to Prob 1349 Maximum Student Taking Exam | dp-with-bitmask-similar-to-prob-1349-max-s654 | Consider each column, since it has much less valid states as compared to each row.\nThe below algorithm can also be extended to work for more than 3 colors scen | finback | NORMAL | 2022-02-15T04:00:11.050004+00:00 | 2022-02-16T00:55:49.717974+00:00 | 172 | false | Consider each column, since it has much less valid states as compared to each row.\nThe below algorithm can also be extended to work for more than 3 colors scenario\n\n```\nclass Solution { \n long[] mask;\n int idx = 0;\n // RRR => 001001001, RGR => 001010001, RGB => 001010100\n public int colorTheGrid(... | 0 | 0 | ['Dynamic Programming', 'Bitmask'] | 0 |
painting-a-grid-with-three-different-colors | c++ | dp | graph | c-dp-graph-by-srv-er-pgcl | \nclass Solution {\npublic:\n int md=1000000007;\n vector<string> st;\n int dp[50][1001];\n void fill(int m,char prev,string s){\n if(m==0)st | srv-er | NORMAL | 2022-02-15T01:19:25.558893+00:00 | 2022-02-15T01:19:25.558927+00:00 | 258 | false | ```\nclass Solution {\npublic:\n int md=1000000007;\n vector<string> st;\n int dp[50][1001];\n void fill(int m,char prev,string s){\n if(m==0)st.push_back(s);\n else{\n string t="RGB";\n for(auto&i:t){\n if(i!=prev) fill(m-1,i,s+i);\n } ... | 0 | 0 | ['Dynamic Programming', 'Graph', 'C'] | 0 |
painting-a-grid-with-three-different-colors | JAVA Intuitive Solution | java-intuitive-solution-by-krishna382nit-gtpo | \nclass Solution {\n public int colorTheGrid(int m, int n) {\n Map<String, Integer> map=new HashMap();\n populateMap(map, m,"");\n if(n= | krishna382nitjsr | NORMAL | 2022-02-06T20:47:52.531195+00:00 | 2022-02-06T20:48:20.154271+00:00 | 425 | false | ```\nclass Solution {\n public int colorTheGrid(int m, int n) {\n Map<String, Integer> map=new HashMap();\n populateMap(map, m,"");\n if(n==1){\n return map.size();\n }\n Map<String, List<String>> map1=new HashMap();\n for(String key:map.keySet()){\n po... | 0 | 0 | [] | 0 |
painting-a-grid-with-three-different-colors | Golang DP solution | golang-dp-solution-by-tjucoder-7fcy | go\nfunc colorTheGrid(m int, n int) int {\n\tbase := make([]int, 0, m)\n\tfor i := 1; i <= 3; i++ {\n\t\tbase = append(base, i)\n\t}\n\tfor i := 2; i <= m; i++ | tjucoder | NORMAL | 2022-01-19T16:02:53.608334+00:00 | 2022-01-19T16:02:53.608365+00:00 | 119 | false | ```go\nfunc colorTheGrid(m int, n int) int {\n\tbase := make([]int, 0, m)\n\tfor i := 1; i <= 3; i++ {\n\t\tbase = append(base, i)\n\t}\n\tfor i := 2; i <= m; i++ {\n\t\tnewBase := make([]int, 0, 64)\n\t\tfor _, v := range base {\n\t\t\tfor j := 1; j <= 3; j++ {\n\t\t\t\tif j == v%10 {\n\t\t\t\t\tcontinue\n\t\t\t\t}\n\... | 0 | 0 | ['Dynamic Programming', 'Go'] | 0 |
painting-a-grid-with-three-different-colors | Java Dp | java-dp-by-prathihaspodduturi-na4b | \nclass Solution {\n public void trav(int m,List<String>list,String s)\n {\n int len=s.length();\n if(len>1 && s.charAt(len-1)==s.charAt(len | prathihaspodduturi | NORMAL | 2021-12-15T12:04:25.634355+00:00 | 2021-12-15T12:04:25.634400+00:00 | 386 | false | ```\nclass Solution {\n public void trav(int m,List<String>list,String s)\n {\n int len=s.length();\n if(len>1 && s.charAt(len-1)==s.charAt(len-2))\n return;\n if(m==0)\n {\n list.add(s);\n return;\n }\n trav(m-1,list,s+"r");\n trav... | 0 | 0 | [] | 0 |
painting-a-grid-with-three-different-colors | [JavaScript] DP with Bit masking | javascript-dp-with-bit-masking-by-tmohan-x1yq | Time O(n * (2^m) * (2^m))\nSpace O(n * 2^(2 * m))\n\nvar colorTheGrid = function (m, n) {\n let dp = Array.from({ length: n }, () => Array(1 << (2 * m)));\n\ | tmohan | NORMAL | 2021-09-15T16:23:18.950901+00:00 | 2021-09-15T16:23:18.950933+00:00 | 130 | false | **Time O(n * (2^m) * (2^m))\nSpace O(n * 2^(2 * m))**\n```\nvar colorTheGrid = function (m, n) {\n let dp = Array.from({ length: n }, () => Array(1 << (2 * m)));\n\n const mod = 1e9 + 7;\n\n const dfs = (rowIdx, colIdx, prevRowColor) => {\n if (rowIdx === n) return 1;\n\n if (colIdx === 0 && dp[r... | 0 | 0 | [] | 0 |
painting-a-grid-with-three-different-colors | Easy DP Approach using UnorderedMap | easy-dp-approach-using-unorderedmap-by-o-oh2l | \n#define ll long long int\nconst int mod=1e9+7;\nclass Solution {\n int r,c;\n string curr;\n vector<string> comb; // all possible valid combinations | ojha1111pk | NORMAL | 2021-09-12T16:30:30.160709+00:00 | 2021-09-12T16:30:30.160763+00:00 | 319 | false | ```\n#define ll long long int\nconst int mod=1e9+7;\nclass Solution {\n int r,c;\n string curr;\n vector<string> comb; // all possible valid combinations to color a column\n unordered_map<string,ll> dp[1003];\npublic:\n\t// finds all possible color combinations to color a given column\n void findCombinat... | 0 | 0 | ['Dynamic Programming', 'C', 'C++'] | 0 |
largest-rectangle-in-histogram | 5ms O(n) Java solution explained (beats 96%) | 5ms-on-java-solution-explained-beats-96-skgtx | For any bar i the maximum rectangle is of width r - l - 1 where r - is the last coordinate of the bar to the right with height h[r] >= h[i] and l - is the last | anton4 | NORMAL | 2016-03-05T12:14:03+00:00 | 2018-10-21T01:39:43.212995+00:00 | 177,468 | false | For any bar `i` the maximum rectangle is of width `r - l - 1` where r - is the last coordinate of the bar to the **right** with height `h[r] >= h[i]` and l - is the last coordinate of the bar to the **left** which height `h[l] >= h[i]`\n\nSo if for any `i` coordinate we know his utmost higher (or of the same height) ne... | 1,357 | 12 | ['Dynamic Programming', 'Java'] | 125 |
largest-rectangle-in-histogram | AC Python clean solution using stack 76ms | ac-python-clean-solution-using-stack-76m-nmv1 | def largestRectangleArea(self, height):\n height.append(0)\n stack = [-1]\n ans = 0\n for i in xrange(len(height)):\n whi | dietpepsi | NORMAL | 2015-10-23T16:45:25+00:00 | 2018-10-24T11:02:29.968831+00:00 | 74,538 | false | def largestRectangleArea(self, height):\n height.append(0)\n stack = [-1]\n ans = 0\n for i in xrange(len(height)):\n while height[i] < height[stack[-1]]:\n h = height[stack.pop()]\n w = i - stack[-1] - 1\n ans = max(ans, h * w)\n ... | 646 | 6 | ['Python'] | 60 |
largest-rectangle-in-histogram | Short and Clean O(n) stack based JAVA solution | short-and-clean-on-stack-based-java-solu-s9u8 | For explanation, please see https://bit.ly/2we8Wfx\n\n\n public int largestRectangleArea(int[] heights) {\n int len = heights.length;\n Stack<I | legendaryengineer | NORMAL | 2015-01-19T20:51:19+00:00 | 2020-04-30T19:30:40.520578+00:00 | 120,429 | false | For explanation, please see https://bit.ly/2we8Wfx\n\n```\n public int largestRectangleArea(int[] heights) {\n int len = heights.length;\n Stack<Integer> s = new Stack<>();\n int maxArea = 0;\n for (int i = 0; i <= len; i++){\n int h = (i == len ? 0 : heights[i]);\n ... | 369 | 12 | ['Java'] | 66 |
largest-rectangle-in-histogram | Video Explanation | video-explanation-by-niits-vacy | Solution Video\n\nhttps://youtu.be/WTbLeCP8rwM\n\n### \u2B50\uFE0F\u2B50\uFE0F Don\'t forget to subscribe to my channel! \u2B50\uFE0F\u2B50\uFE0F\n\n\u25A0 Subs | niits | NORMAL | 2024-06-27T14:29:21.596748+00:00 | 2024-08-17T12:42:16.409320+00:00 | 22,922 | false | # Solution Video\n\nhttps://youtu.be/WTbLeCP8rwM\n\n### \u2B50\uFE0F\u2B50\uFE0F Don\'t forget to subscribe to my channel! \u2B50\uFE0F\u2B50\uFE0F\n\n**\u25A0 Subscribe URL**\nhttp://www.youtube.com/channel/UC9RMNwYTL3SXCP6ShLWVFww?sub_confirmation=1\n\nSubscribers: 6,001\nThank you for your support!\n\n---\n\nhttps:/... | 290 | 0 | ['C++', 'Java', 'Python3', 'JavaScript'] | 3 |
largest-rectangle-in-histogram | My concise C++ solution, AC 90 ms | my-concise-c-solution-ac-90-ms-by-sipipr-vlbv | I push a sentinel node back into the end of height to make the code logic more concise.\n \n\n class Solution {\n public:\n int largestR | sipiprotoss5 | NORMAL | 2014-10-09T01:18:19+00:00 | 2018-10-23T06:25:22.498889+00:00 | 89,874 | false | I push a sentinel node back into the end of height to make the code logic more concise.\n \n\n class Solution {\n public:\n int largestRectangleArea(vector<int> &height) {\n \n int ret = 0;\n height.push_back(0);\n vector<int> index;\n ... | 278 | 10 | [] | 39 |
largest-rectangle-in-histogram | [Java/C++] Explanation going from Brute to Optimal Approach | javac-explanation-going-from-brute-to-op-5m0n | \nWhat the Question is saying, \nGiven an array of integers heights representing the histogram\'s bar height where the width of each bar is 1, \nreturn the area | hi-malik | NORMAL | 2022-01-29T05:17:20.078713+00:00 | 2022-08-03T11:57:13.226306+00:00 | 22,120 | false | ```\nWhat the Question is saying, \nGiven an array of integers heights representing the histogram\'s bar height where the width of each bar is 1, \nreturn the area of the largest rectangle in the histogram.\n```\n\nUnderstand this problem with an example;\n**Input:** `heights = [2,1,5,6,2,3]`\n Solution | segment-tree-solution-just-another-idea-4xuan | Inspired by this solution:\nhttp://www.geeksforgeeks.org/largest-rectangular-area-in-a-histogram-set-1/\n\nJust wanna to provide another idea of how to solve th | xiaohui5319 | NORMAL | 2016-05-20T05:40:14+00:00 | 2016-05-20T05:40:14+00:00 | 31,481 | false | Inspired by this solution:\nhttp://www.geeksforgeeks.org/largest-rectangular-area-in-a-histogram-set-1/\n\nJust wanna to provide another idea of how to solve this question in O(N*logN) theoretically, 40 ms solution:\n\n // Largest Rectangle in Histogram\n // Stack solution, O(NlogN) solution\n \n class SegT... | 191 | 0 | ['C++'] | 112 |
largest-rectangle-in-histogram | AC clean Java solution using stack | ac-clean-java-solution-using-stack-by-je-e0le | public int largestRectangleArea(int[] h) {\n int n = h.length, i = 0, max = 0;\n \n Stack<Integer> s = new Stack<>();\n \n while (i | jeantimex | NORMAL | 2015-09-13T23:10:32+00:00 | 2018-10-01T15:36:57.399300+00:00 | 21,543 | false | public int largestRectangleArea(int[] h) {\n int n = h.length, i = 0, max = 0;\n \n Stack<Integer> s = new Stack<>();\n \n while (i < n) {\n // as long as the current bar is shorter than the last one in the stack\n // we keep popping out the stack and calculate the area ba... | 190 | 5 | ['Java'] | 12 |
largest-rectangle-in-histogram | Simple Divide and Conquer AC solution without Segment Tree | simple-divide-and-conquer-ac-solution-wi-9het | The idea is simple: for a given range of bars, the maximum area can either from left or right half of the bars, or from the area containing the middle two bars. | gildorwang | NORMAL | 2015-01-17T12:40:03+00:00 | 2018-08-29T06:19:37.340555+00:00 | 23,801 | false | The idea is simple: for a given range of bars, the maximum area can either from left or right half of the bars, or from the area containing the middle two bars. For the last condition, expanding from the middle two bars to find a maximum area is `O(n)`, which makes a typical Divide and Conquer solution with `T(n) = 2T(... | 144 | 1 | ['Divide and Conquer', 'C++'] | 15 |
largest-rectangle-in-histogram | Python solution without using stack. (with explanation) | python-solution-without-using-stack-with-sb7b | The algorithm using stack actually is not quite intuitive. At least I won't think about using it at first time.\n\nIt is more natural to think about this way. L | samuri | NORMAL | 2016-10-04T15:21:36.608000+00:00 | 2018-08-26T23:38:45.618608+00:00 | 11,670 | false | The algorithm using stack actually is not quite intuitive. At least I won't think about using it at first time.\n\nIt is more natural to think about this way. Let left[i] to indicate how many bars to the left (including the bar at index i) are equal or higher than bar[i], right[i] is that to the right of bar[i], so the... | 117 | 2 | [] | 24 |
largest-rectangle-in-histogram | 44ms Easy Solution | C++ | 44ms-easy-solution-c-by-giriteja94495-suau | This method is not very intuitive ..i did this with some previous knowledge of similar problems\n\nclass Solution {\npublic:\n int largestRectangleArea(vecto | giriteja94495 | NORMAL | 2020-07-12T15:29:57.187789+00:00 | 2020-07-12T15:32:05.174240+00:00 | 12,091 | false | This method is not very intuitive ..i did this with some previous knowledge of similar problems\n```\nclass Solution {\npublic:\n int largestRectangleArea(vector<int>& heights) {\n // this problem should be solved using stack .\n /* whenever you see a monotonic increase in the input which \n\t\two... | 100 | 4 | ['Stack', 'C', 'C++'] | 11 |
largest-rectangle-in-histogram | ✔️ [Python3] MONOTONIC STACK t(-_-t), Explained | python3-monotonic-stack-t-_-t-explained-jd0a8 | UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.\n\nThe idea is to use a monotonic stack. We iterate over bars a | artod | NORMAL | 2022-01-29T03:35:24.745577+00:00 | 2022-01-29T09:01:31.260792+00:00 | 11,526 | false | **UPVOTE if you like (\uD83C\uDF38\u25E0\u203F\u25E0), If you have any question, feel free to ask.**\n\nThe idea is to use a monotonic stack. We iterate over bars and add them to the stack as long as the last element in the stack is less than the current bar. When the condition doesn\'t hold, we start to calculate area... | 90 | 5 | ['Monotonic Stack', 'Python3'] | 13 |
largest-rectangle-in-histogram | All 4 approaches shown || O(N^2) || O(NlogN) || O(N)(two pass) || O(N)(one pass) with explanation | all-4-approaches-shown-on2-onlogn-ontwo-cnl6w | I will try to explain all the three approaches I know for solving this problem : \n1) Brute Force O(n^2)(too slow TLE)\nwe can compute area of every subarray a | sharan_17 | NORMAL | 2022-01-29T08:21:30.900015+00:00 | 2022-01-29T08:22:08.092344+00:00 | 9,048 | false | I will try to explain all the three approaches I know for solving this problem : \n**1) Brute Force O(n^2)**(too slow **TLE**)\nwe can compute area of every subarray and return the max of them all.\n```\nclass Solution {\npublic:\n int largestRectangleArea(vector<int>& heights) {\n int n = heights.size() ; \... | 81 | 2 | ['Divide and Conquer', 'Stack', 'Tree', 'Recursion', 'C'] | 3 |
largest-rectangle-in-histogram | 📊 Monotonique Stack Solution Intuition (Javascript) | monotonique-stack-solution-intuition-jav-ehn8 | Oh, hard problem it is, isn\'t it? \nLet\'s develop some intuition on how to access it.\n\n## Increasing \nWhen the heights are increasing, we can always build | avakatushka | NORMAL | 2021-08-28T10:20:59.953224+00:00 | 2021-08-28T10:24:12.689723+00:00 | 3,829 | false | Oh, **hard problem** it is, isn\'t it? \nLet\'s develop some intuition on how to access it.\n\n## Increasing \nWhen the heights are **increasing**, we can always build rectangle from a point there a previous height started, to the current point. \n), one trick used | my-concise-code-20ms-stack-based-on-one-o622x | The idea is simple, use a stack to save the index of each vector entry in a ascending order; once the current entry is smaller than the one with the index s.top | lejas | NORMAL | 2015-05-21T01:59:00+00:00 | 2018-10-08T20:08:44.116083+00:00 | 15,181 | false | The idea is simple, use a stack to save the index of each vector entry in a ascending order; once the current entry is smaller than the one with the index s.top(), then that means the rectangle with the height height[s.top()] ends at the current position, so calculate its area and update the maximum. \nThe only trick I... | 55 | 2 | [] | 7 |
largest-rectangle-in-histogram | [C++] Simple Solution | Brute Force ▶️ Optimal | W/ Explanation | c-simple-solution-brute-force-optimal-w-7oax0 | APPROACH 1 : BRUTE FORCE\n\n A Brute force approach would be to one by one consider all bars as starting points and calculate area of all rectangles starting wi | Mythri_Kaulwar | NORMAL | 2022-01-29T02:54:46.629012+00:00 | 2022-01-29T17:31:53.720248+00:00 | 5,796 | false | **APPROACH 1 : BRUTE FORCE**\n\n* A Brute force approach would be to one by one consider all bars as starting points and calculate area of all rectangles starting with every bar. \n* Finally return maximum of all possible areas. \n\n**Time complexity :** O(n^2)\n\n* We can use Divide and Conquer to solve this in O(nLog... | 48 | 4 | ['Stack', 'C', 'C++'] | 3 |
largest-rectangle-in-histogram | 🔥BEATS 💯 % 🎯 |✨SUPER EASY BEGINNERS 👏 | beats-super-easy-beginners-by-codewithsp-83a9 | \n\n---\n\n\n---\n# Intuition\nThe problem can be visualized as calculating the largest rectangle that can be formed in a histogram. The intuition is to use a s | CodeWithSparsh | NORMAL | 2024-12-08T15:53:06.109952+00:00 | 2024-12-08T16:32:23.281582+00:00 | 5,178 | false | \n\n---\n\n\n---\n# Intuition\nThe problem can be visualized as calculating the largest... | 40 | 0 | ['Array', 'Stack', 'C', 'Monotonic Stack', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'Dart'] | 2 |
largest-rectangle-in-histogram | 6 Similar questions || Prev smaller - Next smaller || Monotonic stack || C++ | 6-similar-questions-prev-smaller-next-sm-0na3 | 1. Largest Rectangle in a Histogram: https://leetcode.com/problems/largest-rectangle-in-histogram/\n\n\nclass Solution {\npublic:\n int largestRectangleArea( | hyperVJ | NORMAL | 2021-11-30T07:38:37.092811+00:00 | 2021-11-30T07:38:37.092839+00:00 | 2,824 | false | **1. Largest Rectangle in a Histogram**: https://leetcode.com/problems/largest-rectangle-in-histogram/\n\n```\nclass Solution {\npublic:\n int largestRectangleArea(vector<int>& heights) {\n int n = heights.size();\n vector<int> nsl(n),psl(n);\n \n stack<int> s;\n \n for(int ... | 39 | 0 | ['C', 'Monotonic Stack'] | 4 |
largest-rectangle-in-histogram | [Python] Monotone Increasing Stack, Similar problems attached | python-monotone-increasing-stack-similar-iivn | \nclass Solution:\n def largestRectangleArea(self, heights: List[int]) -> int:\n ## RC ##\n ## APPROACH : MONOTONOUS INCREASING STACK ##\n | 101leetcode | NORMAL | 2020-06-15T10:55:23.819071+00:00 | 2020-06-15T10:56:07.575536+00:00 | 6,365 | false | ```\nclass Solution:\n def largestRectangleArea(self, heights: List[int]) -> int:\n ## RC ##\n ## APPROACH : MONOTONOUS INCREASING STACK ##\n ## Similar to Leetcode: 1475. Final Prices With a Special Discount in a Shop ##\n ## Similar to Leetcode: 907. Sum Of Subarray Minimums ##\n ... | 39 | 0 | ['Python', 'Python3'] | 6 |
largest-rectangle-in-histogram | C++ solution, clean code | c-solution-clean-code-by-lchen77-j4x7 | int largestRectangleArea(vector<int>& height) {\n height.push_back(0);\n const int size_h = height.size();\n stack<int> stk;\n int i | lchen77 | NORMAL | 2015-08-09T19:53:20+00:00 | 2015-08-09T19:53:20+00:00 | 8,215 | false | int largestRectangleArea(vector<int>& height) {\n height.push_back(0);\n const int size_h = height.size();\n stack<int> stk;\n int i = 0, max_a = 0;\n while (i < size_h) {\n if (stk.empty() || height[i] >= height[stk.top()]) stk.push(i++);\n else {\n ... | 38 | 2 | [] | 7 |
largest-rectangle-in-histogram | ✅C++ 100% || Worst to Best approaches with explanation || Easy to understand. | c-100-worst-to-best-approaches-with-expl-6px4 | Read the below solutions to understand the logic.\n\nPlease upvote if you like it\n\nSolution 1: Brute Force Approach (TLE)\n\nIntuition: The intuition behind t | pranjal9424 | NORMAL | 2022-09-15T11:48:43.784943+00:00 | 2022-09-15T16:37:23.034216+00:00 | 3,918 | false | **Read the below solutions to understand the logic.**\n\n***Please upvote if you like it***\n\n**Solution 1: Brute Force Approach (TLE)**\n\n**Intuition:** The intuition behind the approach is taking different bars and finding the maximum width possible using the bar.\n\n {\n if (height==null) return 0;//Shoul... | 33 | 2 | ['Java'] | 7 |
largest-rectangle-in-histogram | ✔ C++ || 💯 Explanation using Stack || O(N) time, space | c-explanation-using-stack-on-time-space-6jg4z | Problem:\n- We are given an array of integers which represent the height of bars in a histogram.\n- The width of each bar is 1.\n- We need to form a rectangle w | GodOfm4a1 | NORMAL | 2023-04-20T06:45:02.441183+00:00 | 2023-04-20T06:47:15.047370+00:00 | 2,150 | false | Problem:\n- We are given an array of integers which represent the height of bars in a histogram.\n- The width of each bar is 1.\n- We need to form a rectangle with these bars with the largest area.\n\nExample :\n\n✅ | Python (Step by step explanation) | beats-99-o-n-python-step-by-step-explana-0oah | Intuition\nThe problem involves finding the largest rectangle that can be formed in a histogram of different heights. The approach is to iteratively process the | monster0Freason | NORMAL | 2023-10-16T17:09:51.462734+00:00 | 2023-11-05T20:28:52.990267+00:00 | 3,439 | false | # Intuition\nThe problem involves finding the largest rectangle that can be formed in a histogram of different heights. The approach is to iteratively process the histogram\'s bars while keeping track of the maximum area found so far.\n\n# Approach\n1. Initialize a variable `maxArea` to store the maximum area found, an... | 27 | 0 | ['Stack', 'Python3'] | 3 |
largest-rectangle-in-histogram | Javascript using stack (Detail breakdown) | javascript-using-stack-detail-breakdown-ilmb1 | Do I explain this good enough?``\njs\n/**\n * @param {number[]} heights\n * @return {number}\n */\nconst largestRectangleArea = function (heights) {\n let ma | harikirito | NORMAL | 2020-03-03T07:31:04.095360+00:00 | 2020-03-03T07:34:13.945507+00:00 | 2,729 | false | Do I explain this good enough?``\n```js\n/**\n * @param {number[]} heights\n * @return {number}\n */\nconst largestRectangleArea = function (heights) {\n let maxArea = 0;\n const stack = [];\n // Append shadow rectangle (height = 0) to both side\n heights = [0].concat(heights).concat([0]);\n for (let i =... | 26 | 0 | ['JavaScript'] | 6 |
largest-rectangle-in-histogram | O(n) stack c++ solution 12ms 中文详细解释 | on-stack-c-solution-12ms-zhong-wen-xiang-s5wb | \u601D\u8DEF\u662F\n\n\u56E0\u4E3A\u76F8\u90BB\u7684\u6BD4curr\u9AD8\u7684\u5757\u80AF\u5B9A\u53EF\u4EE5\u4F5C\u4E3A \u9AD8\u5EA6\u4E3Aheights[curr]\u7684\u957F | alanzwy | NORMAL | 2019-01-12T07:08:22.886392+00:00 | 2019-01-12T07:08:22.886437+00:00 | 2,866 | false | \u601D\u8DEF\u662F\n\n\u56E0\u4E3A\u76F8\u90BB\u7684\u6BD4curr\u9AD8\u7684\u5757\u80AF\u5B9A\u53EF\u4EE5\u4F5C\u4E3A \u9AD8\u5EA6\u4E3Aheights[curr]\u7684\u957F\u65B9\u5F62\u7684\u4E00\u90E8\u5206\n\u6240\u4EE5\u89C2\u5BDF\u5230\u9AD8\u5EA6\u4E3Aheights[curr]\u7684\u957F\u65B9\u5F62\u7684\u6700\u5927\u9762\u79EF\u662F ... | 23 | 3 | [] | 4 |
largest-rectangle-in-histogram | C++ | Stack | Next Smaller Element | c-stack-next-smaller-element-by-reetisha-bbtx | \nclass Solution {\npublic:\n vector<int> nextSmallerRight(vector<int> heights, int n){\n vector<int> res(n);\n stack<int> st;\n \n | reetisharma | NORMAL | 2021-10-07T13:30:04.364779+00:00 | 2021-10-24T17:50:23.039312+00:00 | 2,022 | false | ```\nclass Solution {\npublic:\n vector<int> nextSmallerRight(vector<int> heights, int n){\n vector<int> res(n);\n stack<int> st;\n \n for(int i=n-1; i>=0; i--){\n while(!st.empty() && heights[st.top()]>=heights[i])\n st.pop();\n if(st.empty())\n ... | 20 | 1 | ['Stack', 'C'] | 4 |
largest-rectangle-in-histogram | C++/Java/Python/JavaScript || ✅🚀 Stack || ✔️🔥TC : O(n) | cjavapythonjavascript-stack-tc-on-by-dev-mbni | Intuition:\nThe intuition behind the solution is to use two arrays, nsr (nearest smaller to the right) and nsl (nearest smaller to the left), to determine the b | devanshupatel | NORMAL | 2023-05-13T13:52:30.722382+00:00 | 2023-05-13T13:52:30.722413+00:00 | 5,194 | false | # Intuition:\nThe intuition behind the solution is to use two arrays, `nsr` (nearest smaller to the right) and `nsl` (nearest smaller to the left), to determine the boundaries of the rectangle for each bar in the histogram. By calculating the area for each rectangle and keeping track of the maximum area, we can find th... | 19 | 0 | ['Monotonic Stack', 'Python', 'C++', 'Java', 'JavaScript'] | 1 |
largest-rectangle-in-histogram | Java | TC: O(N) | SC: O(N) | Optimal Stack solution | java-tc-on-sc-on-optimal-stack-solution-tkik8 | java\n/**\n * Using stack to save the increasing height index.\n *\n * Time Complexity: O(N) --> Each element is visited maximum twice. (Once pushed\n * in stac | NarutoBaryonMode | NORMAL | 2021-10-13T10:00:52.891549+00:00 | 2021-10-13T10:09:07.735125+00:00 | 2,316 | false | ```java\n/**\n * Using stack to save the increasing height index.\n *\n * Time Complexity: O(N) --> Each element is visited maximum twice. (Once pushed\n * in stack and once popped for stack)\n *\n * Space Complexity: O(N)\n *\n * N = Length of the input array.\n */\nclass Solution {\n public int largestRectangleAre... | 18 | 0 | ['Array', 'Stack', 'Java'] | 3 |
largest-rectangle-in-histogram | Share my 2ms Java solution. Beats 100% Java submissions | share-my-2ms-java-solution-beats-100-jav-tivk | public class Solution {\n public int largestRectangleArea(int[] heights) {\n if (heights == null || heights.length == 0) return 0;\n | ya_ah | NORMAL | 2016-03-13T19:38:49+00:00 | 2016-03-13T19:38:49+00:00 | 6,010 | false | public class Solution {\n public int largestRectangleArea(int[] heights) {\n if (heights == null || heights.length == 0) return 0;\n return getMax(heights, 0, heights.length);\n } \n int getMax(int[] heights, int s, int e) {\n if (s + 1 >= e) return heights[s... | 17 | 2 | ['Java'] | 9 |
largest-rectangle-in-histogram | Solution Using Stack(C++) | solution-using-stackc-by-surabhi1997mitr-gc9x | This problem is particularly almost similar to Trapping Rainwater Problem.\nHere in this post I would like to highlight both their solutions so one could dop a | surabhi1997mitra | NORMAL | 2021-04-04T15:33:22.036940+00:00 | 2021-04-04T15:33:22.036974+00:00 | 1,717 | false | This problem is particularly almost similar to Trapping Rainwater Problem.\nHere in this post I would like to highlight both their solutions so one could dop a comparitive study of the approach and figure out their individual details.\n\n**Trapping Rainwater Problem (Approach Using Stack)**\n\n```\nint trap(vector<int... | 15 | 3 | ['Stack', 'C'] | 0 |
largest-rectangle-in-histogram | ✔️ 100% Fastest Swift Solution, time: O(n), space: O(n). | 100-fastest-swift-solution-time-on-space-8gro | \nclass Solution {\n // - Complexity:\n // - time: O(n), where n is the length of heights.\n // - space: O(n), where n is the length of heights.\n\ | sergeyleschev | NORMAL | 2022-04-07T05:44:40.681628+00:00 | 2022-04-07T05:44:40.681678+00:00 | 642 | false | ```\nclass Solution {\n // - Complexity:\n // - time: O(n), where n is the length of heights.\n // - space: O(n), where n is the length of heights.\n\n func largestRectangleArea(_ heights: [Int]) -> Int {\n var stack = [Int]()\n let n = heights.count\n var ans = 0\n \n ... | 14 | 0 | ['Swift'] | 0 |
largest-rectangle-in-histogram | Stack C++ - Easy to understand (180ms) [2 Approaches] | stack-c-easy-to-understand-180ms-2-appro-gnft | Here are two approaches for the problem:\n1. We use previous smaller element and next smaller element and compare those two to \n\tget our desired max rectangle | NeerajSati | NORMAL | 2022-01-29T04:31:26.395379+00:00 | 2022-01-29T05:26:16.505051+00:00 | 1,226 | false | Here are two approaches for the problem:\n1. We use previous smaller element and next smaller element and compare those two to \n\tget our desired max rectangle area.\n\t\n\tWe have to use two separate functions here to find the previous smaller and next smaller element.\n\tThe algorithm to make it happen is this:\n\t`... | 14 | 0 | ['Stack', 'C'] | 2 |
largest-rectangle-in-histogram | Python Solution Using Monotonic Stack With Explanation | python-solution-using-monotonic-stack-wi-hfaj | Problem Description\n\nThe problem requires finding the area of the largest rectangle that can be formed in a histogram represented by an array of non-negative | KhadimHussainDev | NORMAL | 2024-04-13T07:29:36.314301+00:00 | 2024-04-14T11:40:50.533084+00:00 | 2,856 | false | ## Problem Description\n\nThe problem requires finding the area of the largest rectangle that can be formed in a histogram represented by an array of non-negative integers `heights`, where each bar\'s width is 1 unit.\n\n## Approach Explanation\n\nThis solution uses a stack-based approach to calculate the maximum recta... | 13 | 1 | ['Stack', 'Monotonic Stack', 'Python3'] | 4 |
largest-rectangle-in-histogram | My C++ DP solution, 16ms, easy to understand! | my-c-dp-solution-16ms-easy-to-understand-z7k8 | I think it might be O(n), or very close to O(n), it cannot be O(n^2).\nCan anybody help to prove?\n\n\n int largestRectangleArea(vector& height) {\n i | phu1ku | NORMAL | 2015-07-21T09:57:29+00:00 | 2015-07-21T09:57:29+00:00 | 4,734 | false | I think it might be O(n), or very close to O(n), it cannot be O(n^2).\nCan anybody help to prove?\n\n\n int largestRectangleArea(vector<int>& height) {\n int n = height.size(), ans = 0, p;\n vector<int> left(n,0), right(n,n);\n for (int i = 1;i < n;++i) {\n p = i-1;\n while... | 13 | 0 | ['C++'] | 7 |
largest-rectangle-in-histogram | Detailed explanation of this problem O(n) | detailed-explanation-of-this-problem-on-7xs0a | Idea 1:\n\tDivide and Conquer using segment tree search. Time & space O(nlgn)\n\thttp://www.geeksforgeeks.org/largest-rectangular-area-in-a-histogram-set-1/\n\t | zehua2 | NORMAL | 2016-08-19T23:54:49.344000+00:00 | 2016-08-19T23:54:49.344000+00:00 | 1,525 | false | Idea 1:\n\tDivide and Conquer using segment tree search. Time & space O(nlgn)\n\thttp://www.geeksforgeeks.org/largest-rectangular-area-in-a-histogram-set-1/\n\t\n\t\nIdea 2:\n\tFor each bar, we calculate maximum area with current bar as the smallest one in the rectangle. If we want to maximize the area of this rectangl... | 13 | 0 | [] | 2 |
largest-rectangle-in-histogram | Largest Rectangle in Histogram - Easy Approach explained with images | largest-rectangle-in-histogram-easy-appr-lqd0 | Problem Statement Explained\n Describe your first thoughts on how to solve this problem. \nThis question wants to find the largest rectangle (ie largest area) t | ak9807 | NORMAL | 2024-07-14T11:48:17.047177+00:00 | 2024-07-15T14:16:21.751520+00:00 | 3,137 | false | # Problem Statement Explained\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis question wants to find the largest rectangle (ie largest area) that can be formed in the given histogram.\nWe are given an array of heights of the histogram bars and width of all is supposed to be 1 unit.\n\n---\n\n... | 12 | 0 | ['Array', 'Stack', 'Monotonic Stack', 'Python', 'Java'] | 1 |
largest-rectangle-in-histogram | [JAVA] Clean Code, O(N) Solution Using Stack Data Structure | java-clean-code-on-solution-using-stack-wy9b9 | \nclass Solution {\n public int largestRectangleArea(int[] heights) {\n \n int maximumRectangle = 0;\n Stack<Integer> stack = new Stack< | anii_agrawal | NORMAL | 2020-09-24T19:05:43.440043+00:00 | 2021-04-28T19:35:10.095185+00:00 | 1,003 | false | ```\nclass Solution {\n public int largestRectangleArea(int[] heights) {\n \n int maximumRectangle = 0;\n Stack<Integer> stack = new Stack<> ();\n \n for (int i = 0; i <= heights.length; i++) {\n while (!stack.isEmpty () && (i == heights.length || heights[stack.peek ()] ... | 12 | 1 | ['Stack', 'Java'] | 1 |
largest-rectangle-in-histogram | faster than 88.25% [easy-understanding][c++][stack] | faster-than-8825-easy-understandingcstac-xy68 | \n class Solution {\n public:\n int largestRectangleArea(vector& heights) {\n stack s;\n in | rajat_gupta_ | NORMAL | 2020-08-20T06:47:31.067136+00:00 | 2020-08-20T07:02:14.060119+00:00 | 1,179 | false | \n class Solution {\n public:\n int largestRectangleArea(vector<int>& heights) {\n stack<int> s;\n int maxArea = 0;\n int area = 0;\n int i;\n for(i=0; i < heights.size();){\n ... | 12 | 1 | ['Stack', 'C', 'C++'] | 0 |
largest-rectangle-in-histogram | Java solution with explanations in Chinese | java-solution-with-explanations-in-chine-npra | S1:\u53CC\u91CD\u904D\u5386\u6CD5\n\n\u672C\u9898\u8981\u6C42\u7684\u662F\u4E00\u6BB5\u8FDE\u7EED\u7684\u77E9\u5F62\uFF0C\u80FD\u591F\u7EC4\u6210\u7684\u9762\u7 | wuruoye | NORMAL | 2019-01-25T06:56:33.228332+00:00 | 2019-01-25T06:56:33.228376+00:00 | 947 | false | S1:\u53CC\u91CD\u904D\u5386\u6CD5\n\n\u672C\u9898\u8981\u6C42\u7684\u662F\u4E00\u6BB5\u8FDE\u7EED\u7684\u77E9\u5F62\uFF0C\u80FD\u591F\u7EC4\u6210\u7684\u9762\u79EF\u6700\u5927\u7684\u77E9\u5F62\u7684\u9762\u79EF\uFF0C\u6240\u4EE5\uFF0C\u53EA\u8981\u80FD\u591F\u6C42\u51FA\u8FD9\u4E00\u6BB5\u77E9\u5F62\u7684\u4F4D\u7F6E\... | 12 | 3 | [] | 6 |
largest-rectangle-in-histogram | [Python] Stack Solution O(N) Time - using approach of Next Smaller to Left and Right | python-stack-solution-on-time-using-appr-p92u | \nclass Solution:\n def largestRectangleArea(self, heights: List[int]) -> int:\n n = len(heights)\n # left boundary => next smaller element to | SamirPaulb | NORMAL | 2022-05-28T09:38:51.882768+00:00 | 2022-05-28T09:42:14.487702+00:00 | 2,418 | false | ```\nclass Solution:\n def largestRectangleArea(self, heights: List[int]) -> int:\n n = len(heights)\n # left boundary => next smaller element to left\n stack = []\n nextSmallerLeft = [0]*n\n for i in range(n):\n while stack and heights[stack[-1]] >= heights[i]:\n ... | 11 | 0 | ['Stack', 'Monotonic Stack', 'Python', 'Python3'] | 3 |
largest-rectangle-in-histogram | [C++] O(n) Stack solution || Explaination with diagram | c-on-stack-solution-explaination-with-di-9en6 | Approach:\nArea of reactangle is height * width\nSince we want to maximise the area , we can either increase height or width. But since the heights are already | iShubhamRana | NORMAL | 2022-01-29T05:36:00.087881+00:00 | 2022-01-29T05:38:30.909711+00:00 | 629 | false | **Approach:**\nArea of reactangle is height * width\nSince we want to maximise the area , we can either increase height or width. But since the heights are already fixed , we need to find the maximum width for each height,\n\nThe naive approach to implement this is to traverse and for each height traverse to left and r... | 11 | 0 | ['Monotonic Stack'] | 0 |
largest-rectangle-in-histogram | ✅C++ solution using Stack & Arrays with full explanations | c-solution-using-stack-arrays-with-full-0j6vx | If you\u2019re interested in coding you can join my Discord Server, link in the comment section. Also if you find any mistake please let me know. Thank you!\u27 | dhruba-datta | NORMAL | 2022-01-22T13:06:46.186991+00:00 | 2022-01-22T17:45:09.380765+00:00 | 760 | false | > **If you\u2019re interested in coding you can join my Discord Server, link in the comment section. Also if you find any mistake please let me know. Thank you!\u2764\uFE0F**\n> \n\n---\n## Explanation:\n\n### Solution 01\n\n- Here we\u2019ll take a stack & 2 arrays to solve it.\n- We\u2019ll find the left and right sm... | 11 | 0 | ['C', 'C++'] | 1 |
largest-rectangle-in-histogram | Python solution | python-solution-by-zitaowang-fu9r | Initialize an empty stack and iterate over the elements in heights. Suppose we are at index i, we can construct stack in such a way that it contains all the rec | zitaowang | NORMAL | 2021-01-19T04:07:32.835863+00:00 | 2021-01-19T04:08:24.256004+00:00 | 846 | false | Initialize an empty `stack` and iterate over the elements in `heights`. Suppose we are at index `i`, we can construct `stack` in such a way that it contains all the rectangles that started at an index `j <= i`, and can survive `i` (i.e., `height[j] <= height[i]`). By construction, the rectangles in `stack` will have a ... | 11 | 0 | [] | 1 |
largest-rectangle-in-histogram | TLE for test case 3000 consecutive 1s | tle-for-test-case-3000-consecutive-1s-by-ef9q | It passes the custom testcase in 24ms, but when I submit it's TLE..\n\nCode:\n\n public class Solution {\n public int largestRectangleArea(int[] heigh | arkansol | NORMAL | 2015-10-19T03:09:56+00:00 | 2015-10-19T03:09:56+00:00 | 1,445 | false | It passes the custom testcase in 24ms, but when I submit it's TLE..\n\nCode:\n\n public class Solution {\n public int largestRectangleArea(int[] height) {\n Stack<Integer> maxIdx = new Stack<Integer>();\n int area = 0;\n for (int i = 0; i < height.length; i++) {\n ... | 11 | 0 | [] | 2 |
largest-rectangle-in-histogram | Java Easy Solution | java-easy-solution-by-suryanshhbtu-1pko | \nclass Solution {\n public int largestRectangleArea(int[] heights) {\n int ans=0;\n int ps[]=previousSmall(heights);\n int ns[]=nextSma | suryanshhbtu | NORMAL | 2022-02-05T19:20:22.559218+00:00 | 2022-02-05T19:20:22.559258+00:00 | 949 | false | ```\nclass Solution {\n public int largestRectangleArea(int[] heights) {\n int ans=0;\n int ps[]=previousSmall(heights);\n int ns[]=nextSmall(heights);\n for(int i=0 ;i<heights.length ;i++){\n ans=Math.max(ans,(ns[i]-ps[i]-1)*heights[i]);\n }\n return ans;\n }\... | 10 | 0 | ['Stack', 'Java'] | 0 |
largest-rectangle-in-histogram | Easy C++ sol with just one traversal | easy-c-sol-with-just-one-traversal-by-by-cazx | Pre-requisite = Previous smaller element using stack.\n\nIdea :-The idea is to use the concept of next smaller element and previous smaller element, we will mai | byteZorvin | NORMAL | 2022-01-29T07:58:51.097926+00:00 | 2022-01-29T09:22:58.099945+00:00 | 301 | false | **Pre-requisite** = Previous smaller element using stack.\n\n**Idea** :-The idea is to use the concept of next smaller element and previous smaller element, we will maintain the index of previous smaller element in the stack and we can get the next smaller element when we pop the element from the stack as the element i... | 10 | 0 | [] | 1 |
largest-rectangle-in-histogram | python using stack | python-using-stack-by-texasroh-0fcz | \nclass Solution:\n def largestRectangleArea(self, heights: List[int]) -> int:\n ans = 0\n stack = []\n heights = [0] + heights + [0]\n | texasroh | NORMAL | 2020-02-16T06:44:15.079848+00:00 | 2020-02-16T06:44:15.079884+00:00 | 1,729 | false | ```\nclass Solution:\n def largestRectangleArea(self, heights: List[int]) -> int:\n ans = 0\n stack = []\n heights = [0] + heights + [0]\n for i in range(len(heights)):\n while(stack and heights[stack[-1]] > heights[i]):\n j = stack.pop()\n ans = m... | 10 | 0 | ['Stack', 'Python'] | 2 |
largest-rectangle-in-histogram | 16ms [c++] 10-line code with stack | 16ms-c-10-line-code-with-stack-by-cq2159-e18b | int largestRectangleArea(vector<int>& height) {\n height.push_back(0); \n int len = height.size(),res = 0, cur=1;\n int s[len+1]={0};\n | cq2159 | NORMAL | 2015-10-03T20:35:45+00:00 | 2015-10-03T20:35:45+00:00 | 3,221 | false | int largestRectangleArea(vector<int>& height) {\n height.push_back(0); \n int len = height.size(),res = 0, cur=1;\n int s[len+1]={0};\n s[0]=-1;\n for(int i=1;i<len;i++){\n while(cur && height[i]<height[s[cur]])\n res = max(res, height[s[cur]] * (i-s[--cu... | 10 | 1 | ['C++'] | 3 |
largest-rectangle-in-histogram | 10 line c++ solution | 10-line-c-solution-by-kenigma-wbkg | \n\n class Solution {\n public:\n int largestRectangleArea(vector& heights) {\n int res = 0;\n for (int i = 0; i < heights.si | kenigma | NORMAL | 2016-03-03T18:10:05+00:00 | 2018-09-12T13:20:56.131499+00:00 | 2,515 | false | \n\n class Solution {\n public:\n int largestRectangleArea(vector<int>& heights) {\n int res = 0;\n for (int i = 0; i < heights.size(); ++i) {\n if (i + 1 < heights.size() && heights[i] <= heights[i+1]) continue; // find the local max (greater than left and right)\n ... | 10 | 1 | ['C++'] | 3 |
largest-rectangle-in-histogram | ✅ Simple clean stack approach with explanation | simple-clean-stack-approach-with-explana-0mx2 | Approach\n1. Monotonic Stack: Utilize a monotonic stack to keep track of heights and indices in ascending order. If the height encountered is less than the heig | MostafaAE | NORMAL | 2024-03-24T03:16:47.119682+00:00 | 2024-03-24T03:16:47.119710+00:00 | 1,185 | false | # Approach\n1. Monotonic Stack: Utilize a monotonic stack to keep track of heights and indices in ascending order. If the height encountered is less than the height at the top of the stack, pop elements from the stack and calculate the area of the rectangles until the current index.\n2. Compute Largest Area: While popp... | 9 | 0 | ['Stack', 'Monotonic Stack', 'C++'] | 1 |
largest-rectangle-in-histogram | Easy understanding solution with monotonic stack C++ (image explanation included) | easy-understanding-solution-with-monoton-zk9g | Intuition\n Describe your first thoughts on how to solve this problem. \nTry to find the longest width w.r.t. height[i], which means we have to find the most le | sunskyxh | NORMAL | 2023-12-21T01:28:30.861537+00:00 | 2023-12-21T01:30:30.396734+00:00 | 663 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTry to find the longest width w.r.t. `height[i]`, which means we have to find the most left (called `start`) and right (called `end`) positions where the all the heights between `[start, end]` are greater or equal to `height[i]`.\n\n# App... | 9 | 0 | ['Greedy', 'Monotonic Stack', 'C++'] | 0 |
largest-rectangle-in-histogram | Simplest C# solution with stack<int> | simplest-c-solution-with-stackint-by-sto-brey | \n\npublic class Solution {\n public int LargestRectangleArea(int[] heights) {\n int n = heights.Length;\n int max = 0;\n var stack = ne | stodorov | NORMAL | 2023-09-23T12:04:13.772142+00:00 | 2023-09-23T12:04:13.772161+00:00 | 591 | false | \n```\npublic class Solution {\n public int LargestRectangleArea(int[] heights) {\n int n = heights.Length;\n int max = 0;\n var stack = new Stack<int>();\n\n for(int i = 0; i <= n; i++)\n {\n var height = i < n ? heights[i] : 0;\n\n while(stack.Any() && heigh... | 9 | 0 | ['C#'] | 1 |
largest-rectangle-in-histogram | C++ || Stack Solution || Next smaller to left &Right | c-stack-solution-next-smaller-to-left-ri-2bi0 | \t// find the index vector for next smaller to left\n\t// find the index vector for next smaller to right \n\t// width will be right[i]-left[i]-1 \n\t// here st | anubhavsingh11 | NORMAL | 2022-06-24T17:55:28.808295+00:00 | 2022-06-24T17:55:28.808340+00:00 | 465 | false | \t// find the index vector for next smaller to left\n\t// find the index vector for next smaller to right \n\t// width will be right[i]-left[i]-1 \n\t// here stack<pair<int,int>> first element in pair denotes height[i] and \n\t// second element of pair corresponds to index i \n\n\tclass Solution {\n\tpublic:\n\t\t// ne... | 9 | 0 | ['Stack', 'C'] | 1 |
largest-rectangle-in-histogram | JavaScript Using Stack Beat 97% | javascript-using-stack-beat-97-by-contro-3zwu | Time Complexity = O(N)\nSpace Complexity = O(N)\njavascript\nvar largestRectangleArea = function(heights) {\n // to deal with last element without going out | control_the_narrative | NORMAL | 2020-07-16T15:57:07.544484+00:00 | 2020-07-16T18:15:11.288860+00:00 | 1,562 | false | Time Complexity = O(N)\nSpace Complexity = O(N)\n```javascript\nvar largestRectangleArea = function(heights) {\n // to deal with last element without going out of bound\n heights.push(0)\n const stack = [];\n let maxArea = 0, curr, currH, top, topH, area;\n \n for(let i = 0; i < heights.length; i++) {... | 9 | 1 | ['JavaScript'] | 2 |
largest-rectangle-in-histogram | Short C++ solution use stack AC 28ms | short-c-solution-use-stack-ac-28ms-by-cc-lb52 | class Solution {\n public:\n int largestRectangleArea(vector<int>& height) {\n height.push_back(0);\n int result=0;\n | ccweikui | NORMAL | 2015-06-06T04:45:44+00:00 | 2018-08-15T22:53:27.420024+00:00 | 2,362 | false | class Solution {\n public:\n int largestRectangleArea(vector<int>& height) {\n height.push_back(0);\n int result=0;\n stack<int> indexStack;\n for(int i=0;i<height.size();i++){\n while(!indexStack.empty()&&height[i]<height[indexStack.top()]){\n ... | 9 | 0 | [] | 1 |
largest-rectangle-in-histogram | Explination of the stack solution (python solution) | explination-of-the-stack-solution-python-t8qk | To get the bigest rectangle area, we should check every rectangle with lowest point at i=1...n. \n\nDefine Si := the square with lowest point at i. \nTo calcu | jddymx | NORMAL | 2015-10-09T20:40:21+00:00 | 2015-10-09T20:40:21+00:00 | 2,804 | false | To get the bigest rectangle area, we should check every rectangle with lowest point at i=1...n. \n\nDefine `Si := the square with lowest point at i`. \nTo calculate faster this `Si` , we have to use a stack `stk` which stores some *indices*. \n\nThe elements in stk satisfy these **properties**: \n\n1. the indices a... | 9 | 0 | ['Python'] | 1 |
largest-rectangle-in-histogram | Python solution with detailed explanation | python-solution-with-detailed-explanatio-b89x | Solution with discussion https://discuss.leetcode.com/topic/77574/python-solution-with-detailed-explanation\n\nLargest Rectangle in Histogram https://leetcode.c | gabbu | NORMAL | 2017-02-02T17:34:57.476000+00:00 | 2018-08-27T02:01:09.537966+00:00 | 2,204 | false | **Solution with discussion** https://discuss.leetcode.com/topic/77574/python-solution-with-detailed-explanation\n\n**Largest Rectangle in Histogram** https://leetcode.com/problems/largest-rectangle-in-histogram/\n\nhttp://www.geeksforgeeks.org/largest-rectangle-under-histogram/\nhttp://www.geeksforgeeks.org/largest-rec... | 9 | 1 | [] | 3 |
largest-rectangle-in-histogram | My 16ms C++ O(n) code without a stack | my-16ms-c-on-code-without-a-stack-by-bea-fqrm | Basically, I find the range (left[i], right[i]) within which the height of every element is at least height[i]. For each i, there is a rectangular with area hei | bear2015 | NORMAL | 2015-09-05T16:35:55+00:00 | 2015-09-05T16:35:55+00:00 | 2,534 | false | Basically, I find the range (left[i], right[i]) within which the height of every element is at least height[i]. For each i, there is a rectangular with area height[i] * (right[i]-left[i]+1), and the largest area must be one of them. Then run all i to find the largest area. Please notice it takes O(n) time to fill left[... | 9 | 0 | ['Dynamic Programming', 'C++'] | 2 |
largest-rectangle-in-histogram | TWO APPROACHES USING LEFT AND RIGHT MINIMUM || SINGLE PASS || O( N ) ✅✅ | two-approaches-using-left-and-right-mini-n6ad | Approach 1\n\n### Intuition:\nWe can solve this problem using a stack to efficiently calculate the left and right boundaries of each bar.\n\n### Approach:\n1. U | Sabari_Raj_004 | NORMAL | 2024-05-07T06:36:31.386832+00:00 | 2024-05-07T06:36:31.386865+00:00 | 1,004 | false | ## Approach 1\n\n### Intuition:\nWe can solve this problem using a stack to efficiently calculate the left and right boundaries of each bar.\n\n### Approach:\n1. Use two passes:\n - In the first pass, calculate the left boundary of each bar.\n - In the second pass, calculate the right boundary of each bar and simul... | 8 | 0 | ['Array', 'Stack', 'Monotonic Stack', 'Java'] | 1 |
largest-rectangle-in-histogram | ✅Easy✨||C++ || Beats 100% || With proper Explanation | easyc-beats-100-with-proper-explanation-g5zhb | Intuition\n Describe your first thoughts on how to solve this problem. \nWe will use two function next smaller and prev smaller element function and by getting | olakade33 | NORMAL | 2024-02-22T14:08:02.718362+00:00 | 2024-02-22T14:08:02.718396+00:00 | 1,199 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe will use two function next smaller and prev smaller element function and by getting that value we will find width and length is properly the height of histogram by simply just multiplying width and length we will get area and at the en... | 8 | 0 | ['C++'] | 0 |
largest-rectangle-in-histogram | Shortest Code || Rat bhi skte ho | shortest-code-rat-bhi-skte-ho-by-surajma-l0vc | daalo -1 stack mein, aur 0 ko nums ke back mein\n2. jaldi se krlo pura yaad, interviewer ke saamne krdena anuvaad\n3. upvote kroge to milegi jrur, ldki nhi jo | surajmamgai | NORMAL | 2022-09-01T22:33:28.529643+00:00 | 2022-09-01T22:46:35.415497+00:00 | 424 | false | 1. **daalo -1 stack mein, aur 0 ko nums ke back mein**\n2. **jaldi se krlo pura yaad, interviewer ke saamne krdena anuvaad**\n3. **upvote kroge to milegi jrur, ldki nhi job huzur**\n\n\n\n\n```\n int largestRectangleArea(vector<int>& nums) {\n stack<int> st;\n st.push(-1);\n nums.push_back(0);\n int ans=0... | 8 | 0 | [] | 1 |
largest-rectangle-in-histogram | Detailed explanation with diagrams and examples using pen and paper|| Brute-force and Optimized | detailed-explanation-with-diagrams-and-e-3tks | Explanation\n\n\n\n\n\n\n\nJAVA Code\n1. Brute force (To understand he concept)\n\nclass Solution {\n public int largestRectangleArea(int[] heights) {\n | rachana_raut | NORMAL | 2022-01-29T17:03:19.883695+00:00 | 2022-01-29T17:03:19.883727+00:00 | 478 | false | **Explanation**\n\n\n\n\n\n\n\n**JAVA Code**\n1. Brute force (To understand he concept)\n```\nclas... | 8 | 0 | ['Java'] | 1 |
largest-rectangle-in-histogram | Easy Brute Force approach | C++ | easy-brute-force-approach-c-by-vaishnavi-p870 | \nclass Solution {\npublic:\n int largestRectangleArea(vector<int>& heights) {\n int n=heights.size(),prev,next;\n long long maxi=0;\n f | vaishnavi_vv | NORMAL | 2022-01-29T12:10:01.750753+00:00 | 2022-01-29T12:10:01.750798+00:00 | 384 | false | ```\nclass Solution {\npublic:\n int largestRectangleArea(vector<int>& heights) {\n int n=heights.size(),prev,next;\n long long maxi=0;\n for(int i=0;i<n;i++)\n {\n int c=1;\n prev=i-1;\n next=i+1;\n while(prev>=0 && heights[i]<=heights[prev]) /... | 8 | 1 | ['C'] | 2 |
largest-rectangle-in-histogram | Python3 O(N log(N)) divide and conquer approach explained | python3-on-logn-divide-and-conquer-appro-34wq | The first approach I came up with was to try every possible pair of start and end positions, keeping track of the minimum encountered height since the correspon | alexeir | NORMAL | 2020-04-04T10:59:16.952373+00:00 | 2020-04-04T14:58:34.490542+00:00 | 1,005 | false | The first approach I came up with was to try every possible pair of start and end positions, keeping track of the minimum encountered height since the corresponding starting position and pick the pair achieving maximum area. This can be implemented in O(N^2) time complexity. This approach exceeded the time limit (at le... | 8 | 0 | ['Divide and Conquer', 'Python3'] | 1 |
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