question_slug stringlengths 3 77 | title stringlengths 1 183 | slug stringlengths 12 45 | summary stringlengths 1 160 ⌀ | author stringlengths 2 30 | certification stringclasses 2
values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
classes | content stringlengths 4 576k | upvotes int64 0 11.5k | downvotes int64 0 358 | tags stringlengths 2 193 | comments int64 0 2.56k |
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number-of-ways-to-split-array | One Line Solution | one-line-solution-by-charnavoki-eaj1 | null | charnavoki | NORMAL | 2025-01-03T07:09:38.601727+00:00 | 2025-01-03T07:09:38.601727+00:00 | 19 | false |
```javascript []
const waysToSplitArray = ([a, ...nums], s = nums.reduce((x, y) => y + x)) =>
nums.reduce((w, v) => ([w, a, s] = [w + (a >= s), a + v, s - v])[0], 0);
``` | 2 | 0 | ['JavaScript'] | 0 |
number-of-ways-to-split-array | Step-by-Step Solution: Number of Ways to Split Array (100% Easy, 0ms) | step-by-step-solution-number-of-ways-to-8vc9g | IntuitionThe problem is about finding the number of ways to split an array into two non-empty parts such that the sum of the first part is greater than or equal | user1233nm | NORMAL | 2025-01-03T07:02:49.077530+00:00 | 2025-01-04T05:51:41.615688+00:00 | 26 | false | # Intuition
The problem is about finding the number of ways to split an array into two non-empty parts such that the sum of the first part is greater than or equal to the sum of the second part. The **prefix sum technique** provides an efficient way to calculate the cumulative sum and perform quick range queries.
# Ap... | 2 | 0 | ['Array', 'Prefix Sum', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'JavaScript'] | 1 |
number-of-ways-to-split-array | Beats 100% Easy C++ Using Prefix and Suffix -- Ways to Split Array into Two Parts | beats-100-easy-c-using-prefix-and-suffix-c9lb | Method 1 -- Optimized way of the second method given after thisIntuitionThe task is to split the array into two non-empty parts such that the sum of elements in | Abhishek-Verma01 | NORMAL | 2025-01-03T06:40:32.480841+00:00 | 2025-01-03T06:40:32.480841+00:00 | 27 | false | # Method 1 -- Optimized way of the second method given after this
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The task is to split the array into two non-empty parts such that the sum of elements in the left part `(lSum)` is greater than or equal to the sum of elements in the right p... | 2 | 0 | ['Array', 'Sliding Window', 'Suffix Array', 'Prefix Sum', 'C++'] | 0 |
number-of-ways-to-split-array | Simple prefixSum solution✅✅ | simple-prefixsum-solution-by-_shubham_22-67y1 | IntuitionThe task is to find the number of ways to split an array nums nums such that:
Both parts are non-empty.
The sum of the left part (prefixSum) is greater | _shubham_22 | NORMAL | 2025-01-03T06:11:34.107285+00:00 | 2025-01-03T06:11:34.107285+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The task is to find the number of ways to split an array nums nums such that:
1. Both parts are non-empty.
1. The sum of the left part (prefixSum) is greater than or equal to the sum of the right part (suffixSum).
By observing:
- prefixSum... | 2 | 0 | ['Array', 'Prefix Sum', 'C++'] | 0 |
number-of-ways-to-split-array | 🌟 Beat 100% | 🔥Optimized Approach 💡 | Beginner-Friendly 🐣 | Easy Solution ✅ | beat-100-optimized-approach-beginner-fri-62ra | Intuition 🧠
The goal is to find splits in the array such that the sum of the left part is greater than or equal to the sum of the right part. This involves calc | shubhamrajpoot_ | NORMAL | 2025-01-03T05:45:59.828286+00:00 | 2025-01-03T05:45:59.828286+00:00 | 52 | false | # Intuition 🧠
- The goal is to find splits in the array such that the sum of the left part is greater than or equal to the sum of the right part. This involves calculating prefix sums efficiently.
# *Brute Force Approach* 🚶♂️
## *Approach*:
- Compute two prefix sum arrays (left and right) for the left and right pa... | 2 | 0 | ['Prefix Sum', 'Python', 'C++', 'Java', 'Go', 'Rust', 'Ruby', 'JavaScript', 'C#', 'Dart'] | 0 |
number-of-ways-to-split-array | 2 ms🚀|| 100% Beats ✅ || JAVA || Simple Logic🌟 | 2-ms-100-beats-java-simple-logic-by-siva-p7w0 | IntuitionApproachComplexity
Time complexity: O(n + (n-1))
Space complexity:
Code | Sivaram11 | NORMAL | 2025-01-03T05:29:23.074767+00:00 | 2025-01-03T05:29:23.074767+00:00 | 27 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: O... | 2 | 0 | ['Java'] | 1 |
number-of-ways-to-split-array | 🔥 Super Clean 4-Line C++ Solution | One-Pass Magic! | super-clean-4-line-c-solution-one-pass-m-2kr8 | IntuitionThe problem requires determining the number of valid ways to split the array such that the sum of elements on the left side is greater than or equal to | sachin1604 | NORMAL | 2025-01-03T05:24:59.669482+00:00 | 2025-01-03T05:24:59.669482+00:00 | 16 | false | # Intuition
The problem requires determining the number of valid ways to split the array such that the sum of elements on the left side is greater than or equal to the sum of elements on the right side. The key insight is that we can compute the prefix sum and suffix sum efficiently by traversing the array once and upd... | 2 | 0 | ['Array', 'Prefix Sum', 'C++'] | 1 |
number-of-ways-to-split-array | Easy Solution || 2ms ✅ || Beats 100%👌|| Prefix Sum 👍 | easy-solution-2ms-beats-100-prefix-sum-b-fzwu | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Sorna_Prabhakaran | NORMAL | 2025-01-03T05:22:12.752428+00:00 | 2025-01-03T05:22:12.752428+00:00 | 15 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<... | 2 | 0 | ['Java'] | 0 |
number-of-ways-to-split-array | Java and C++, Easy Solution | java-and-c-easy-solution-by-divyaaroraa-3lx7 | IntuitionThe problem asks for the number of valid splits of an array, where the sum of the left part is greater than or equal to the sum of the right part. This | divyaaroraa | NORMAL | 2025-01-03T05:18:36.918706+00:00 | 2025-01-03T05:18:36.918706+00:00 | 11 | false | # Intuition
The problem asks for the number of valid splits of an array, where the sum of the left part is greater than or equal to the sum of the right part. This can be efficiently solved using prefix sums.
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
1. First, calculate the total s... | 2 | 0 | ['C++'] | 0 |
number-of-ways-to-split-array | Beats 100% users in JAVA | beats-100-users-in-java-by-muskaan_1801-akyv | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | muskaan_1801 | NORMAL | 2025-01-03T05:12:29.978830+00:00 | 2025-01-03T05:12:29.978830+00:00 | 16 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 2 | 0 | ['Java'] | 0 |
number-of-ways-to-split-array | Counting Valid Splits in an Array Based on Sum Comparison | counting-valid-splits-in-an-array-based-oen8t | IntuitionThe problem asks us to determine the number of valid splits in the array such that the sum of the left subarray is greater than or equal to the sum of | jasurbek_fm | NORMAL | 2025-01-03T05:05:34.492787+00:00 | 2025-01-03T05:05:34.492787+00:00 | 5 | false | # Intuition
The problem asks us to determine the number of valid splits in the array such that the sum of the left subarray is greater than or equal to the sum of the right subarray. A straightforward approach involves computing prefix sums and comparing them for each potential split.
# Approach
**1. Calculate Total S... | 2 | 0 | ['Prefix Sum', 'C++'] | 0 |
number-of-ways-to-split-array | C++ solution using prefix sum and suffix sum. stay permission less | c-solution-using-prefix-sum-and-suffix-s-3yho | IntuitionApproachsimply create the prefix sum array and suffix sum array and then
if sum of prefix is grreater than equal to the next index of suffix sum then p | Gufrankhan | NORMAL | 2025-01-03T03:55:41.019573+00:00 | 2025-01-03T03:55:41.019573+00:00 | 17 | false | # Intuition
# Approach
simply create the prefix sum array and suffix sum array and then
if sum of prefix is grreater than equal to the next index of suffix sum then possible to split the array we count the number of ways and return it;
# Complexity
- Time complexity:
0(N) // size of nums
- Space complex... | 2 | 0 | ['C++'] | 0 |
number-of-ways-to-split-array | Number of Ways to Split Array- Two Approaches✅|| Beats 100% 🚀|| Prefix Sum & Single Pass || JAVA ☕ | number-of-ways-to-split-array-two-approa-5hoh | IntuitionThe problem asks us to find the number of valid splits in an array such that the sum of the left part is greater than or equal to the sum of the right | Megha_Mathur18 | NORMAL | 2025-01-03T03:50:33.506618+00:00 | 2025-01-03T03:50:33.506618+00:00 | 20 | false | # Intuition
The problem asks us to find the number of valid splits in an array such that the sum of the left part is greater than or equal to the sum of the right part. My initial thought was to maintain prefix sums for both left and right parts of the array, but I later realized that a single pass with cumulative sums... | 2 | 0 | ['Java'] | 0 |
number-of-ways-to-split-array | clear code with simple looping and tracking | 100% beats | clear-code-with-simple-looping-and-track-xe35 | Approach
Tracking the prefix sum and sufix sum with simple for loop and comparing prefix and sufix.
use long varible for prefix and sufix else we facing the err | keshavakumar_jr | NORMAL | 2025-01-03T03:37:57.567070+00:00 | 2025-01-03T03:37:57.567070+00:00 | 12 | false |
# Approach
<!-- Describe your approach to solving the problem. -->
1. Tracking the prefix sum and sufix sum with simple for loop and comparing prefix and sufix.
2. use long varible for prefix and sufix else we facing the error.
3. upvote if u like my approch and explanation.
# Complexity
- Time complexity:O(n)
<!-- Ad... | 2 | 0 | ['Prefix Sum', 'Java'] | 0 |
number-of-ways-to-split-array | Simplest Code🚀 || Beat 100% || Complete Explanation | simplest-code-beat-100-complete-explanat-3lwn | Upvote if you find this helpful! 👍IntuitionThe problem involves finding the number of valid splits of an array such that the sum of the left part is greater tha | CodeNikET | NORMAL | 2025-01-03T03:31:39.807944+00:00 | 2025-01-03T03:31:39.807944+00:00 | 36 | false | # **Upvote if you find this helpful! 👍**
# Intuition
The problem involves finding the number of valid splits of an array such that the sum of the left part is greater than or equal to the sum of the right part. To achieve this, we can calculate prefix sums and suffix sums for efficient comparisons at each split point... | 2 | 0 | ['Array', 'Math', 'Prefix Sum', 'C++'] | 1 |
number-of-ways-to-split-array | JAVA || 3MS 🚀🚀|| BEATS 60 % || EASY SOLUTION✅✅ || | java-3ms-beats-60-easy-solution-by-kavi_-yqn3 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | kavi_k | NORMAL | 2025-01-03T03:10:30.370528+00:00 | 2025-01-03T03:10:30.370528+00:00 | 18 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 2 | 0 | ['Java'] | 0 |
largest-submatrix-with-rearrangements | [C++] Clean and Clear, With Intuitive Pictures, O(m * n * logn) | c-clean-and-clear-with-intuitive-picture-pcwo | This question is very similar to \n84. Largest Rectangle in Histogram\n85. Maximal Rectangle\n\nThey are both hard questions, so don\'t be frustrated if you can | aincrad-lyu | NORMAL | 2021-01-17T04:15:47.754131+00:00 | 2021-01-17T08:29:19.230247+00:00 | 13,659 | false | This question is very similar to \n[84. Largest Rectangle in Histogram](https://leetcode.com/problems/largest-rectangle-in-histogram/)\n[85. Maximal Rectangle](https://leetcode.com/problems/maximal-rectangle/)\n\nThey are both hard questions, so don\'t be frustrated if you cannot solve it in the contest.\n\nFirst think... | 821 | 19 | [] | 76 |
largest-submatrix-with-rearrangements | Java | 6ms | easy understanding with comments and images | java-6ms-easy-understanding-with-comment-sqoi | This is like this one https://leetcode.com/problems/maximal-rectangle/\nbut since each block can move, we cannot use that method (it will cause TLE).\nMy though | rioto9858 | NORMAL | 2021-01-17T04:10:11.353425+00:00 | 2021-01-17T13:59:57.508304+00:00 | 6,200 | false | This is like this one [https://leetcode.com/problems/maximal-rectangle/](http://)\nbut since each block can move, we cannot use that method (it will cause TLE).\nMy thought is like this : \n1. we need the rectangle for each column so we need to memorize that. --> change the matrix\n2. the columns can move ----> sort... | 206 | 5 | [] | 16 |
largest-submatrix-with-rearrangements | 【Video】Give me 10 minutes - how we think about a solution | video-give-me-10-minutes-how-we-think-ab-b8eb | Intuition\nUsing sort for width.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/odAv92zWKqs\n\nThese days, many people watch my video more than 70% of whole tim | niits | NORMAL | 2023-11-26T07:49:27.916070+00:00 | 2023-11-28T20:00:39.302975+00:00 | 4,215 | false | # Intuition\nUsing sort for width.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/odAv92zWKqs\n\nThese days, many people watch my video more than 70% of whole time, which is good to my channel reputation. Thank you!\n\nI\'m creating this video or post with the goal of ensuring that viewers, whether they read the post o... | 134 | 0 | ['C++', 'Java', 'Python3', 'JavaScript'] | 11 |
largest-submatrix-with-rearrangements | ✅ Beats 100% - Explained with [ Video ] - Simple Sorting - Visualized Too | beats-100-explained-with-video-simple-so-b8ia | \n\n # YouTube Video Explanation:\n\n<a href="https://youtu.be/biPaIwFMISI">https://youtu.be/biPaIwFMISI</a>\n\n**\uD83D\uDD25 Please like, share, and subscribe | lancertech6 | NORMAL | 2023-11-26T03:31:35.574328+00:00 | 2023-11-27T17:23:02.079063+00:00 | 11,200 | false | \n\n<!-- # YouTube Video Explanation:\n\n[https://youtu.be/biPaIwFMISI](https://youtu.be/biPaIwFMISI)\n\n**\uD83D\uDD25 Please like, share, and subscribe to support our channel\'s mis... | 84 | 2 | ['Array', 'Greedy', 'Sorting', 'Matrix', 'Python', 'C++', 'Java', 'JavaScript'] | 13 |
largest-submatrix-with-rearrangements | C++ Solution | Easy Implementation | c-solution-easy-implementation-by-invuln-nbqp | \nint largestSubmatrix(vector<vector<int>>& matrix) {\n \n int i, j, ans = 0, n = matrix.size(), m = matrix[0].size();\n \n for(i = | invulnerable | NORMAL | 2021-01-17T04:00:49.698617+00:00 | 2021-01-17T13:43:38.351537+00:00 | 4,775 | false | ```\nint largestSubmatrix(vector<vector<int>>& matrix) {\n \n int i, j, ans = 0, n = matrix.size(), m = matrix[0].size();\n \n for(i = 0; i < m; i++)\n {\n for(j = 1; j < n; j++)\n {\n if(matrix[j][i] == 1)\n matrix[j][i] = matri... | 82 | 4 | [] | 14 |
largest-submatrix-with-rearrangements | Optimal O(m*n) time, O(n) space, no sort :) | optimal-omn-time-on-space-no-sort-by-ale-lo45 | We first explain an O(m * n * log n) algorithm that, internally, uses an O(n * log n) sort for each row. That algorithm is similar to most other solutions for | alex_salcianu | NORMAL | 2021-01-17T04:43:49.992374+00:00 | 2021-01-18T09:47:08.263615+00:00 | 1,725 | false | We first explain an O(m * n * log n) algorithm that, internally, uses an O(n * log n) sort for each row. That algorithm is similar to most other solutions for this problem. It is good-enough for the current tests, but not optimal. Then, we explain how to remove that sort() and obtain an O(m * n) algorithm. Finally,... | 32 | 0 | [] | 4 |
largest-submatrix-with-rearrangements | Explanation with Picture. Variation of Maximal Rectangle and Largest Rectangle in Histogram | explanation-with-picture-variation-of-ma-s9l3 | This problem is very similar to Maximum rectangular area and Largest Rectangle in Histogram. The only variation here is that we can swap the columns and put the | cut_me_half | NORMAL | 2021-01-17T04:01:00.137554+00:00 | 2021-01-18T04:57:20.524436+00:00 | 2,406 | false | This problem is very similar to Maximum rectangular area and Largest Rectangle in Histogram. The only variation here is that we can swap the columns and put the columns next to each other to optimize the answer.\n\nWell, we can just sort the current row in which we are and then can directly use `largestRectangularArea`... | 31 | 8 | [] | 7 |
largest-submatrix-with-rearrangements | Easy to Read || Beginner friendly || No Fluff || C++/Java/Python | easy-to-read-beginner-friendly-no-fluff-zu9q1 | Treat every column in the matrix as a height bar, where the base of that height represents total number of consecutives ones above it included himself. \n\n\n | Cosmic_Phantom | NORMAL | 2023-11-26T02:43:39.314790+00:00 | 2023-11-26T04:18:22.814244+00:00 | 3,460 | false | Treat every column in the matrix as a height bar, where the base of that height represents total number of consecutives ones above it included himself. \n\n```\n [[0,0,1],\n [1,1,1],\n [1,0,1]]\nheight->[2,0,3] -> sort -> [3,2,0] -> \n```\n\nNow iterate over this sorted row and consider the largest ... | 26 | 0 | ['Array', 'Math', 'Sorting', 'Matrix'] | 5 |
largest-submatrix-with-rearrangements | [Python3] Concise solution | python3-concise-solution-by-kunqian-0-dcvw | Intuition\n\nFor each row, find the histogram height of each column. \nSort it and find the max rectangle you could find(refer to Question 84).\n\n# Complexity\ | kunqian-0 | NORMAL | 2021-01-17T04:03:20.569527+00:00 | 2021-01-17T07:18:56.342083+00:00 | 957 | false | # **Intuition**\n\nFor each row, find the histogram height of each column. \nSort it and find the max rectangle you could find(refer to [Question 84]( https://leetcode.com/problems/largest-rectangle-in-histogram/)).\n<br/>\n# **Complexity**\nTime: `O(m * n * log(n))`\nSpace: `O(n)`\n<br/>\n# **Python3**\n```python\ndef... | 21 | 7 | [] | 2 |
largest-submatrix-with-rearrangements | 🔥Beginner Friendly Solution with Detail Explanation🔥|| 🔥Daily Challenges🔥 | beginner-friendly-solution-with-detail-e-qwae | Intuition\n\n\n# Complexity\n- Time complexity:O(2nm)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity here, e | Shree_Govind_Jee | NORMAL | 2023-11-26T03:46:54.680703+00:00 | 2023-11-26T03:46:54.680725+00:00 | 1,214 | false | # Intuition\n\n\n# Complexity\n- Time complexity:$$O(2*n*m)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\... | 19 | 1 | ['Array', 'Greedy', 'Sorting', 'Matrix', 'Java'] | 5 |
largest-submatrix-with-rearrangements | Javascript | Simple Solution w/ Explanation & Visuals | beats 100% / 100% | javascript-simple-solution-w-explanation-f57b | Idea:\n\nSince the elements in each column can\'t move, we can think of each consecutive group of 1s in each column as a solid pillar within that column.\n\nmat | sgallivan | NORMAL | 2021-01-18T07:31:21.992871+00:00 | 2021-01-18T07:31:41.047778+00:00 | 608 | false | ***Idea:***\n\nSince the elements in each column can\'t move, we can think of each consecutive group of **1**s in each column as a solid pillar within that column.\n```\nmatrix: [ [ 0, 1, 0, 1, 0 ],\n [ 1, 1, 0, 1, 1 ],\n [ 1, 1, 1, 0, 1 ],\n [ 1, 0, 1, 1, 0 ],\n [ 1, 1, 1,... | 17 | 0 | ['JavaScript'] | 3 |
largest-submatrix-with-rearrangements | Simple Python3 | 9 Lines | Beats 100% | Detailed Explanation | simple-python3-9-lines-beats-100-detaile-a0iz | Logic: Prefix sum from top to the bottom row for every column, by doing this we have number of continuous 1st stacked on top so far at matrix[i][j] . So for exa | sushanthsamala | NORMAL | 2021-01-17T04:01:17.262825+00:00 | 2021-01-17T14:15:46.860527+00:00 | 2,169 | false | Logic: Prefix sum from top to the bottom row for every column, by doing this we have number of continuous 1st stacked on top so far at matrix[i][j] . So for example if the matrix is \n[0,0,1]\n[1,1,1]\n[1,0,1]\nWe have the prefix summed matrix:\n[0, 0, 1]\n[1, 1, 2]\n[2, 0, 3]\nNow the we could just go row by row and c... | 17 | 2 | ['Python3'] | 3 |
largest-submatrix-with-rearrangements | Beat 100%! O(m*n*logn ) easy solution | beat-100-omnlogn-easy-solution-by-lawren-47rr | Example 1\n\n[0,0,1],\n[1,1,1],\n[1,0,1] \n\nFor every column, we count the the number of continuous 1 by the current position. the result matrix will be\n\n[0 | lawrenceflag | NORMAL | 2021-01-17T04:01:22.574114+00:00 | 2021-01-17T04:30:14.089900+00:00 | 1,120 | false | Example 1\n\n[0,0,1],\n[1,1,1],\n[1,0,1] \n\nFor every column, we count the the number of continuous 1 by the current position. the result matrix will be\n\n[0,0,1],\n[1,1,2],\n[2,0,3] \n\nNow let\u2019s consider every row. \nFor the first row, we reversely sort the row\nWe got [1,0,0]. It\'s very easy to get the larg... | 16 | 8 | [] | 4 |
largest-submatrix-with-rearrangements | C++ sort row vs Freq count||124 ms Beats 98.30% | c-sort-row-vs-freq-count124-ms-beats-983-4n3q | Intuition\n Describe your first thoughts on how to solve this problem. \nUse the matrix to store the height for each column.\n\nLater try other approach!\n# App | anwendeng | NORMAL | 2023-11-26T03:10:57.628557+00:00 | 2023-11-26T08:34:09.690840+00:00 | 2,575 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse the `matrix` to store the height for each column.\n\nLater try other approach!\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUse sort!\n\nThe approach using counting sort leads to TLE & fails , for the testcase... | 11 | 1 | ['Array', 'Hash Table', 'Sorting', 'Counting Sort', 'C++'] | 3 |
largest-submatrix-with-rearrangements | 【Video】Sorting solution | video-sorting-solution-by-niits-ylte | Intuition\nUsing sort for width.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/odAv92zWKqs\n\nThese days, many people watch my video more than 70% of whole tim | niits | NORMAL | 2024-04-23T16:38:01.350989+00:00 | 2024-04-23T16:38:01.351018+00:00 | 129 | false | # Intuition\nUsing sort for width.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/odAv92zWKqs\n\nThese days, many people watch my video more than 70% of whole time, which is good to my channel reputation. Thank you!\n\nI\'m creating this video or post with the goal of ensuring that viewers, whether they read the post o... | 8 | 0 | ['C++', 'Java', 'Python3', 'JavaScript'] | 0 |
largest-submatrix-with-rearrangements | Beat 100% - Easy solution with explanation similar problem - maximal-rectangle | beat-100-easy-solution-with-explanation-j9phz | intuition\n matrix[i][j] is sum of continous 1\'s\n sort the each row to keep largest matrix because we can swap the columns\n\nadd the continous 1\'s from top | aravind_dev | NORMAL | 2021-01-17T04:03:08.636047+00:00 | 2021-01-17T15:36:20.754502+00:00 | 617 | false | **intuition**\n* matrix[i][j] is sum of continous 1\'s\n* sort the each row to keep largest matrix because we can swap the columns\n\nadd the continous 1\'s from top to bottom.\nsort the each row to find the largest matrix\n\nEx:\n```\nmatrix= [[1,0,0],\n\t\t[1,0,1],\n\t\t[1,1,1]];\n```\n\nafter counting continous 1\'s... | 7 | 0 | [] | 5 |
largest-submatrix-with-rearrangements | [C++] count 1s and then sort | c-count-1s-and-then-sort-by-ddddyliu-7cbf | \nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) {\n int m = matrix.size(), n = matrix[0].size();\n vector<vecto | ddddyliu | NORMAL | 2021-01-17T04:01:56.924656+00:00 | 2021-01-17T04:30:09.288373+00:00 | 804 | false | ```\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) {\n int m = matrix.size(), n = matrix[0].size();\n vector<vector<int>> cntOne(m, vector<int>(n, 0));\n for (int j = 0; j < n; j++) {\n if (matrix[0][j] == 1) {cntOne[0][j] = 1;}\n }\n for ... | 7 | 1 | ['C', 'Sorting'] | 1 |
largest-submatrix-with-rearrangements | 😀 Fully Explained Solution With Example and Dry Run 💡 | fully-explained-solution-with-example-an-nna4 | Problem \nFinding the maximum area of submatrix which has 1 in its each cell.\n\n# How do we calculate area?\nArea = Length * breadth\nIn this problem, we will | mohd-usman | NORMAL | 2023-11-26T15:43:56.004730+00:00 | 2023-11-26T15:43:56.004755+00:00 | 297 | false | # Problem \nFinding the maximum area of submatrix which has 1 in its each cell.\n\n# How do we calculate area?\n**$$Area = Length * breadth$$**\nIn this problem, we will calculate through **width** and **height**.\n\n**But wait, how to find the height \uD83E\uDD14?**\nWell, we have to find the height of those cells tha... | 6 | 2 | ['C++'] | 2 |
largest-submatrix-with-rearrangements | [C++] ✅ Histogram Max Area Finding Problem || Count ones and sort | c-histogram-max-area-finding-problem-cou-zfrs | Approach:\n\nImagine each column in the matrix as a vertical bar, where its height represents the count of consecutive ones above it, including itself.\n\nIt is | anupam8nith | NORMAL | 2023-11-26T09:22:31.151474+00:00 | 2023-11-27T15:16:02.418735+00:00 | 220 | false | **Approach:**\n\nImagine each column in the matrix as a vertical bar, where its height represents the count of consecutive ones above it, including itself.\n\nIt is similar to **Histogram Area Finding Problem:** `Calculating the maximum area of a histogram for each row: This helps in determining the maximum rectangle t... | 6 | 0 | ['C', 'Sorting'] | 2 |
largest-submatrix-with-rearrangements | C++||Largest Submatrix With Rearrangements to Largest Rectangle in Histogram||logic explained | clargest-submatrix-with-rearrangements-t-4wda | Simple conversion from Largest Submatrix With Rearrangements to Largest Rectangle in Histogram\nIdea:\n\n1.for each a[i][j] cell count number of continuos 1\'s | tejpratapp468 | NORMAL | 2021-01-17T04:06:47.385130+00:00 | 2021-01-20T07:20:56.835425+00:00 | 387 | false | **Simple conversion from Largest Submatrix With Rearrangements to Largest Rectangle in Histogram\nIdea:**\n\n1.for each a[i][j] cell count number of continuos 1\'s in its column,so we have column height of a[i][j];\n2.for each row we will sort the row in descending order of column heights since you are allowed to rear... | 6 | 0 | [] | 2 |
largest-submatrix-with-rearrangements | Simple C++ solution || TC: O(nmlog(m)) || SC:O(1) | simple-c-solution-tc-onmlogm-sco1-by-zee-6k7k | Intuition\nThis code finds the largest submatrix with all 1s in a binary matrix. It first transforms the matrix to represent the height of \'buildings\' of 1s. | Zeeshan_2512 | NORMAL | 2023-11-26T02:31:25.136786+00:00 | 2023-11-26T02:31:25.136807+00:00 | 936 | false | # Intuition\nThis code finds the largest submatrix with all 1s in a binary matrix. It first transforms the matrix to represent the height of \'buildings\' of 1s. Then, for each row, it calculates the maximum area of the rectangle that can be formed using the \'building\' at each cell as the shortest \'building\'. The m... | 5 | 0 | ['C++'] | 1 |
largest-submatrix-with-rearrangements | Python solution easy-understanding with comments | python-solution-easy-understanding-with-p2x9c | class Solution:\n\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n row, col = len(matrix), len(matrix[0])\n for j in range(col): | flyingspa | NORMAL | 2021-04-12T09:29:09.989971+00:00 | 2021-04-12T09:29:09.990002+00:00 | 608 | false | class Solution:\n\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n row, col = len(matrix), len(matrix[0])\n for j in range(col): # calculate the prefix consecutive one for each column \n for i in range(1,row):\n if matrix[i][j]:\n matrix[i][j]... | 5 | 0 | ['Python'] | 0 |
largest-submatrix-with-rearrangements | Python solution: Largest Rectangular Area in a Rearrangable Histogram | python-solution-largest-rectangular-area-ib1f | \n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n def largestRectangleArea(heights):\n heights.sort(reverse=True)\n | otoc | NORMAL | 2021-01-17T04:06:22.620016+00:00 | 2021-01-17T04:07:56.980001+00:00 | 402 | false | ```\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n def largestRectangleArea(heights):\n heights.sort(reverse=True)\n res = heights[0]\n for i in range(1, len(heights)):\n res = max(res, heights[i] * (i + 1))\n return res\n \n ... | 5 | 0 | [] | 0 |
largest-submatrix-with-rearrangements | Easy explained with best approach|| C++||greedy | easy-explained-with-best-approach-cgreed-ng4b | Intuition\nIntution is the precomputation of cells find the height for every column in the subsequent array.Then sorting of row according in non decreasing orde | dev_ved | NORMAL | 2023-11-26T06:11:44.526081+00:00 | 2023-11-26T06:11:44.526112+00:00 | 770 | false | # Intuition\nIntution is the precomputation of cells find the height for every column in the subsequent array.Then sorting of row according in non decreasing order and calculating the area since the cell number would provide the height and take width for subsequent column.\n\n# Approach\n1.precomputation\n2. for any ro... | 4 | 0 | ['Math', 'Greedy', 'Sorting', 'Matrix', 'C++'] | 0 |
largest-submatrix-with-rearrangements | ✅ Beats 96.02% 🚀 c++ code 🔥 | beats-9602-c-code-by-omar_walied_ismail-gqz8 | Intuition\n- I need all indexes in result matrix with only ones.\n\n# Approach\n- Save for everyone in the matrix, the number of consecutive ones before it.\n- | omar_walied_ismail | NORMAL | 2023-11-26T05:26:19.032101+00:00 | 2023-11-26T05:26:19.032131+00:00 | 305 | false | # Intuition\n- I need all indexes in result matrix with only ones.\n\n# Approach\n- Save for everyone in the matrix, the number of consecutive ones before it.\n- Take these numbers and sort it in ascending order. \n- Iterate on these numbers from back to front to get the greater high every time.\n- Take the minimum eve... | 4 | 0 | ['Math', 'Dynamic Programming', 'Sorting', 'Matrix', 'C++'] | 3 |
largest-submatrix-with-rearrangements | 🚀😎 BEATS 100% | MOST EASY APPROACH🔥🔥| C++ | JAVA | PYTHON | JS | beats-100-most-easy-approach-c-java-pyth-bn2s | Intuition\nUPVOTE IF YOU FOUND THIS HELPFUL !!\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n\n\n Describe your approach to solvi | The_Eternal_Soul | NORMAL | 2023-11-26T02:48:04.202426+00:00 | 2023-11-26T02:48:04.202459+00:00 | 995 | false | # Intuition\nUPVOTE IF YOU FOUND THIS HELPFUL !!\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n\n\n<!-- Describe your approach to solving the pro... | 4 | 0 | ['Array', 'Greedy', 'C', 'Sorting', 'Matrix', 'Python', 'C++', 'Java', 'JavaScript'] | 2 |
largest-submatrix-with-rearrangements | Walkthrough of the general algorithm | walkthrough-of-the-general-algorithm-by-bluh5 | Just my notes of a walkthrough of the general algorithm people used to solve this question, hope this helps\n\n\n\t/\n * 1. Precalculate the # of consecutiv | qmscorp | NORMAL | 2021-07-14T01:35:33.985759+00:00 | 2021-07-14T01:35:33.985792+00:00 | 283 | false | Just my notes of a walkthrough of the general algorithm people used to solve this question, hope this helps\n\n\n\t/**\n * 1. Precalculate the # of consecutive ones (vertically) in the matrix,\n * --> which we are actually calculating the height of each column/histogram\n *\n * [0,0,1] [0,0,1]\n... | 4 | 0 | [] | 1 |
largest-submatrix-with-rearrangements | [Python3] Faster than 100% - Easy [Explanation + Comments] | python3-faster-than-100-easy-explanation-1sqc | The core logic of my program is :\nFor any cell (i,j) for all cells in the same row store the number of ones starting from that row.\nFor example, matrix = [[0, | mihirrane | NORMAL | 2021-01-17T04:36:26.540213+00:00 | 2021-01-17T19:07:51.760750+00:00 | 485 | false | The core logic of my program is :\nFor any cell (i,j) for all cells in the same row store the number of ones starting from that row.\nFor example, matrix = [[0,0,1],[1,1,1],[1,0,1]]\nCorresponding map (d) = {1: [2,1,2], 2: [1,1], 0: [3]}\nExplanation of map: \nKey corresponds to the row number. \nLet\'s look at row num... | 4 | 0 | ['Python', 'Python3'] | 1 |
largest-submatrix-with-rearrangements | 🚀✅🚀 Beats 101% -1727. Largest Submatrix With Rearrangements | beats-101-1727-largest-submatrix-with-re-dxqz | Example Explanation\nLet\'s break down the process step by step using an example matrix:\n\nExample Matrix:\n\n[\n [0, 0, 1],\n [1, 1, 1],\n [1, 0, 1]\n]\n\nSte | arshbhatia1551 | NORMAL | 2023-11-26T10:28:13.772766+00:00 | 2023-11-26T10:28:13.772799+00:00 | 158 | false | **Example Explanation**\nLet\'s break down the process step by step using an example matrix:\n\n**Example Matrix:**\n```\n[\n [0, 0, 1],\n [1, 1, 1],\n [1, 0, 1]\n]\n```\n**Step 1: Preprocessing**\n```\n[\n [0, 0, 1],\n [1, 1, 2],\n [2, 0, 3]\n]\n```\nFor each cell matrix[i][j] = 1, update it with the height of consecu... | 3 | 0 | ['Array', 'Greedy', 'C', 'Sorting', 'Matrix', 'Python', 'Java', 'JavaScript'] | 0 |
largest-submatrix-with-rearrangements | O(N*M*LogM) | onmlogm-by-3s_akb-3w04 | \n# Code\n\npublic class Solution {\n public int LargestSubmatrix(int[][] matrix) {\n int Y = matrix.Length; //rows count\n int X = matrix[0 | 3S_AKB | NORMAL | 2023-07-31T20:23:12.380373+00:00 | 2023-07-31T20:23:12.380392+00:00 | 47 | false | \n# Code\n```\npublic class Solution {\n public int LargestSubmatrix(int[][] matrix) {\n int Y = matrix.Length; //rows count\n int X = matrix[0].Length; //columns count\n\n //transform matrix to (matrix[y][x] = count of 1s upper (<y))\n for (int x = 0; x < X; x++) \n ... | 3 | 0 | ['C#'] | 1 |
largest-submatrix-with-rearrangements | C++ | Variation of Leetcode 85.Maximal Rectangle | c-variation-of-leetcode-85maximal-rectan-7l5z | \nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& v) {\n int n=v.size();\n int m=v[0].size();\n int res;\n | GeekyBits | NORMAL | 2022-10-10T12:55:49.812259+00:00 | 2022-10-10T12:55:49.812363+00:00 | 469 | false | ```\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& v) {\n int n=v.size();\n int m=v[0].size();\n int res;\n vector<int> vv(m,0);\n for(int i=0;i<n;i++){\n for(int j=0;j<m;j++){\n if(v[i][j]==1){\n vv[j]+=1;\n ... | 3 | 0 | [] | 0 |
largest-submatrix-with-rearrangements | C++ Solution | O(m * n * logn) | c-solution-om-n-logn-by-horcrux903-mzvd | \nclass Solution {\npublic:\n\n\tint largestSubmatrix(vector<vector<int>>& a)\n\t{\n\t\tint ans = 0;\n\t\tint n = a.size(), m = a[0].size();\n\n\t\tint tmp[m];\ | horcrux903 | NORMAL | 2022-01-04T19:01:27.881769+00:00 | 2022-01-04T19:01:27.881814+00:00 | 150 | false | ```\nclass Solution {\npublic:\n\n\tint largestSubmatrix(vector<vector<int>>& a)\n\t{\n\t\tint ans = 0;\n\t\tint n = a.size(), m = a[0].size();\n\n\t\tint tmp[m];\n\t\tmemset(tmp, 0, sizeof(tmp));\n\n\t\tfor (int i = 0; i < n; i++)\n\t\t{\n\t\t\tvector<int> v;\n\n\t\t\tfor (int j = 0; j < m; j++)\n\t\t\t{\n\t\t\t\tif (... | 3 | 0 | ['Sorting'] | 0 |
largest-submatrix-with-rearrangements | Easy C++ solution || commented | easy-c-solution-commented-by-saiteja_bal-j505 | \nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) {\n \n //for every column keep a track of number of consecutive | saiteja_balla0413 | NORMAL | 2021-07-05T16:32:23.619671+00:00 | 2021-07-05T16:32:23.619704+00:00 | 349 | false | ```\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) {\n \n //for every column keep a track of number of consecutive ones upto the index\n int m=matrix.size();\n int n=matrix[0].size();\n for(int j=0;j<n;j++)\n {\n for(int i=0;i<m;i++... | 3 | 0 | ['C'] | 0 |
largest-submatrix-with-rearrangements | Java Priority Queue Solution | java-priority-queue-solution-by-mayank_p-op8i | https://achievementguru.com/leetcode-1727-largest-submatrix-with-rearrangements-java-solution/\n\nclass Solution {\n public int largestSubmatrix(int[][] matr | mayank_pratap | NORMAL | 2021-01-18T08:24:18.300532+00:00 | 2021-01-18T08:25:20.941304+00:00 | 538 | false | https://achievementguru.com/leetcode-1727-largest-submatrix-with-rearrangements-java-solution/\n```\nclass Solution {\n public int largestSubmatrix(int[][] matrix) {\n //Store matrix[i][j]= total number of consecutive 1s along column j ending at index (i,j).\n for(int i=1;i<matrix.length;i++) //Start... | 3 | 0 | ['Heap (Priority Queue)', 'Java'] | 0 |
largest-submatrix-with-rearrangements | My Java Solution with small explanation 1) Count the 1's in column 2) Sort the rows and count | my-java-solution-with-small-explanation-iy26l | \nclass Solution {\n public int largestSubmatrix(int[][] matrix) {\n // count number of 1 in each column and update it\n int row = matrix.lengt | vrohith | NORMAL | 2021-01-18T05:06:40.617611+00:00 | 2021-01-18T05:06:40.617696+00:00 | 380 | false | ```\nclass Solution {\n public int largestSubmatrix(int[][] matrix) {\n // count number of 1 in each column and update it\n int row = matrix.length;\n int col = matrix[0].length;\n int count = 0;\n for (int i=1; i<row; i++) {\n for (int j=0; j<col; j++) {\n ... | 3 | 0 | ['Sorting', 'Matrix', 'Java'] | 1 |
largest-submatrix-with-rearrangements | C# - Sorting by heights to get max area | c-sorting-by-heights-to-get-max-area-by-0mzex | csharp\npublic int LargestSubmatrix(int[][] matrix)\n{\n\tint[] heights = new int[matrix[0].Length];\n\tint result = 0;\n\n\tfor (int row = 0; row < matrix.Leng | christris | NORMAL | 2021-01-18T02:39:14.126668+00:00 | 2021-01-18T02:39:14.126695+00:00 | 94 | false | ```csharp\npublic int LargestSubmatrix(int[][] matrix)\n{\n\tint[] heights = new int[matrix[0].Length];\n\tint result = 0;\n\n\tfor (int row = 0; row < matrix.Length; row++)\n\t{\n\t\tfor (int column = 0; column < matrix[0].Length; column++)\n\t\t{\n\t\t\tif (matrix[row][column] == 0)\n\t\t\t{\n\t\t\t\theights[column] ... | 3 | 0 | [] | 0 |
largest-submatrix-with-rearrangements | 16 line Python solution (998ms) | 16-line-python-solution-998ms-by-15sawye-psck | Intuition\n Describe your first thoughts on how to solve this problem. \nIterate over rows while keeping track of the height of columns of 1\'s above the row. S | 15sawyer | NORMAL | 2023-11-26T20:04:44.982906+00:00 | 2023-11-26T20:46:36.465858+00:00 | 425 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIterate over rows while keeping track of the height of columns of 1\'s above the row. Sort the heights after each iteration and check the area of all possible rectangles that would fit into the graph of heights.\n# Approach\n<!-- Describe... | 2 | 0 | ['Python3'] | 1 |
largest-submatrix-with-rearrangements | ✅☑[C++/Java/Python/JavaScript] || 3 Approaches || EXPLAINED🔥 | cjavapythonjavascript-3-approaches-expla-szh1 | PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n(Also explained in the code)\n\n#### Approach 1(Sort By Height On Each Baseline Row)\n1. Matrix traversal:\ | MarkSPhilip31 | NORMAL | 2023-11-26T19:40:48.424908+00:00 | 2023-11-26T19:40:48.424930+00:00 | 231 | false | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Sort By Height On Each Baseline Row)***\n1. **Matrix traversal:**\n - The code iterates through each row of the matrix.\n1. **Accumulating consecutive ones:**\n - For each non-zero element in the matrix (... | 2 | 0 | ['Array', 'Greedy', 'C', 'Sorting', 'Matrix', 'C++', 'Java', 'Python3', 'JavaScript'] | 1 |
largest-submatrix-with-rearrangements | One line solution. Runtime 100%!!! | one-line-solution-runtime-100-by-xxxxkav-7cnp | \n\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n return max(max(map(mul, sorted(row, reverse = True), count(1))) fo | xxxxkav | NORMAL | 2023-11-26T13:56:13.945467+00:00 | 2023-11-26T13:56:13.945501+00:00 | 98 | false | \n```\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n return max(max... | 2 | 0 | ['Python3'] | 0 |
largest-submatrix-with-rearrangements | C# Solution for Largest Submatrix With Rearrangements Problem | c-solution-for-largest-submatrix-with-re-qveu | Intuition\n Describe your first thoughts on how to solve this problem. \nThis solution maintains a count of consecutive 1s vertically in the input matrix and th | Aman_Raj_Sinha | NORMAL | 2023-11-26T11:32:19.495160+00:00 | 2023-11-26T11:32:19.495189+00:00 | 46 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis solution maintains a count of consecutive 1s vertically in the input matrix and then sorts each row to find the maximum area of the submatrix containing only 1s after rearranging the columns optimally.\n\n# Approach\n<!-- Describe yo... | 2 | 0 | ['C#'] | 0 |
largest-submatrix-with-rearrangements | C++ | Simple Code | Beats 85.23% of users with C++ | c-simple-code-beats-8523-of-users-with-c-hw63 | 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 | Ankita2905 | NORMAL | 2023-11-26T11:11:14.344781+00:00 | 2023-11-26T11:11:14.344808+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. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
largest-submatrix-with-rearrangements | Easy C++ || O(m*n) solution | easy-c-omn-solution-by-jpank1983-b1vk | Intuition\n Describe your first thoughts on how to solve this problem. \nThis is similar problem to Largest Rectangle in Histogram. We just need to find height | jPank1983 | NORMAL | 2023-11-26T10:59:16.385405+00:00 | 2023-11-26T16:00:44.627777+00:00 | 25 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis is similar problem to Largest Rectangle in Histogram. We just need to find height of every column through every row taking every row as base.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe hight of every co... | 2 | 0 | ['C++'] | 0 |
largest-submatrix-with-rearrangements | Beats 95% simple prefix sum like solution | beats-95-simple-prefix-sum-like-solution-l9k3 | Intuition\n Describe your first thoughts on how to solve this problem. \ncount the continuous ones on each column for each row and sort them in descending order | rk_ninjaa__404 | NORMAL | 2023-11-26T10:28:17.733239+00:00 | 2023-11-26T10:28:17.733256+00:00 | 259 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ncount the continuous ones on each column for each row and sort them in descending order so that the first column contains the most number of continuous ones. Sorting so that we can align the ones continuously to make it a submatrix. \n\n*... | 2 | 0 | ['Greedy', 'Sorting', 'Prefix Sum', 'C++'] | 2 |
largest-submatrix-with-rearrangements | Java Beginner Friendly Code | java-beginner-friendly-code-by-ritikranj-dx2g | Explanation\n Describe your first thoughts on how to solve this problem. \nStep 1: Preprocessing (as given is Hint 1)\nStep 2: Sorting Rows (as given in Hint 2) | ritikranjan12 | NORMAL | 2023-11-26T09:16:00.444249+00:00 | 2023-11-26T09:16:00.444272+00:00 | 156 | false | # Explanation\n<!-- Describe your first thoughts on how to solve this problem. -->\nStep 1: Preprocessing (as given is Hint 1)\nStep 2: Sorting Rows (as given in Hint 2)\nStep 3: Calculate Area\nIterate through each row and calculate the area for each cell.\n\n# Complexity\n- Time complexity: O(matrix.length * matrix[0... | 2 | 0 | ['Array', 'Math', 'Sorting', 'Java'] | 1 |
largest-submatrix-with-rearrangements | [C] Sorting | c-sorting-by-leetcodebug-653t | 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 | leetcodebug | NORMAL | 2023-11-26T05:30:14.013214+00:00 | 2023-11-26T05:30:14.014825+00:00 | 101 | 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: $$O(m * nlogn)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(m * n)$$\n<!-- Add your space com... | 2 | 0 | ['Array', 'Greedy', 'C', 'Sorting', 'Matrix'] | 1 |
largest-submatrix-with-rearrangements | Python3 Solution | python3-solution-by-motaharozzaman1996-63xv | \n\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n row=len(matrix)\n col=len(matrix[0])\n ans=0\n | Motaharozzaman1996 | NORMAL | 2023-11-26T03:00:50.325226+00:00 | 2023-11-26T03:00:50.325247+00:00 | 904 | false | \n```\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n row=len(matrix)\n col=len(matrix[0])\n ans=0\n for r in range(1,row):\n for c in range(col):\n matrix[r][c]+=matrix[r-1][c] if matrix[r][c] else 0\n\n for r in range(row... | 2 | 0 | ['Python', 'Python3'] | 0 |
largest-submatrix-with-rearrangements | ✅ Swift: Easy to Understand and Simple Solution | swift-easy-to-understand-and-simple-solu-o52v | Complexity\n- Time complexity: O(mn.log.n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) \ | sohagkumarbiswas | NORMAL | 2023-11-26T02:58:08.454603+00:00 | 2023-11-26T02:58:08.454620+00:00 | 26 | false | # Complexity\n- Time complexity: $$O(mn.log.n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n func largestSubmatrix(_ matrix: [[Int]]) -> Int {\n var ans = 0\n var hist =... | 2 | 0 | ['Array', 'Greedy', 'Swift', 'Sorting', 'Matrix'] | 0 |
largest-submatrix-with-rearrangements | O(m*n*log(n)) Solution || | omnlogn-solution-by-yashmenaria-asju | Complexity\n- Time complexity:O(mnlog(n))\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity here, e.g. O(n) \n# | yashmenaria | NORMAL | 2023-11-26T02:55:51.379392+00:00 | 2023-11-26T02:55:51.379411+00:00 | 16 | false | # Complexity\n- Time complexity:O(m*n*log(n))\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n# Code\n# C++\n```\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) {\n for (int i = 0; i < mat... | 2 | 0 | ['Array', 'Greedy', 'Sorting', 'Matrix', 'C++', 'Java'] | 0 |
largest-submatrix-with-rearrangements | C# || Solution with LINQ | c-solution-with-linq-by-vdmhunter-k3a3 | Code\n\npublic class Solution\n{\n public int LargestSubmatrix(int[][] matrix)\n {\n for (var i = 1; i < matrix.Length; i++)\n for (var | vdmhunter | NORMAL | 2023-11-26T00:15:51.511349+00:00 | 2023-11-26T00:15:51.511375+00:00 | 83 | false | # Code\n```\npublic class Solution\n{\n public int LargestSubmatrix(int[][] matrix)\n {\n for (var i = 1; i < matrix.Length; i++)\n for (var j = 0; j < matrix[0].Length; j++)\n if (matrix[i][j] == 1)\n matrix[i][j] = matrix[i - 1][j] + 1;\n\n return matri... | 2 | 1 | ['C#'] | 0 |
largest-submatrix-with-rearrangements | Simple greedy approach || Not a medium level question probably hard one 🤨 | simple-greedy-approach-not-a-medium-leve-3s21 | \nclass Solution\n{\n public:\n int largestSubmatrix(vector<vector < int>> &mat)\n {\n int m = mat.size();\n int n = mat[ | Tan_B | NORMAL | 2023-04-19T06:34:01.517736+00:00 | 2023-04-19T06:34:01.517779+00:00 | 121 | false | ```\nclass Solution\n{\n public:\n int largestSubmatrix(vector<vector < int>> &mat)\n {\n int m = mat.size();\n int n = mat[0].size();\n vector<int> h(n, 0);//array for counting height at each level.\n vector<int> x;\n int ans = 0;\n for... | 2 | 0 | ['Math', 'Greedy', 'C', 'Sorting'] | 0 |
largest-submatrix-with-rearrangements | C++ using 2 traversals by Sorting | c-using-2-traversals-by-sorting-by-divya-axhg | \nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& mat) {\n int m= mat.size();\n int n= mat[0].size();\n for(int i= | divyanshugupta5901 | NORMAL | 2022-07-16T15:18:10.110353+00:00 | 2022-07-16T15:18:10.110396+00:00 | 373 | false | ```\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& mat) {\n int m= mat.size();\n int n= mat[0].size();\n for(int i=1; i<m ; i++){\n for(int j=0; j<n; j++){\n if(mat[i][j]!=0)\n mat[i][j]+= mat[i-1][j];\n }\n ... | 2 | 0 | ['C'] | 0 |
largest-submatrix-with-rearrangements | Java solution with approach | java-solution-with-approach-by-soumya00-ad17 | Approach:\n\n1. Traverse the matrix column wise\n2. For every column, calculate the continuous running sum and update the array itself. If any cell value is 0, | soumya00 | NORMAL | 2022-07-16T04:57:22.723395+00:00 | 2022-07-16T04:57:22.723452+00:00 | 186 | false | **Approach:**\n\n1. Traverse the matrix column wise\n2. For every column, calculate the continuous running sum and update the array itself. If any cell value is 0, reset the continuous sum to 0\n3. Next traverse the matrix row wise, and sort every row in decsending order\n4. Calculate the maximum area greedily and stor... | 2 | 0 | ['Greedy', 'Sorting', 'Simulation'] | 0 |
largest-submatrix-with-rearrangements | ✅✅Faster || Easy To Understand || C++ Code | faster-easy-to-understand-c-code-by-__kr-vkrf | This Problem is like Maximum Rectangle In Histogram\n\n Using Sorting\n\n Time Complexity :- O(N * M * logM)\n\n Space Complexity :- O(M)*\n\n\nclass Solution { | __KR_SHANU_IITG | NORMAL | 2022-07-15T05:59:26.285714+00:00 | 2022-07-15T05:59:26.285778+00:00 | 301 | false | * ***This Problem is like Maximum Rectangle In Histogram***\n\n* ***Using Sorting***\n\n* ***Time Complexity :- O(N * M * logM)***\n\n* ***Space Complexity :- O(M)***\n\n```\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) {\n \n int n = matrix.size();\n \n i... | 2 | 0 | ['C', 'Sorting', 'C++'] | 0 |
largest-submatrix-with-rearrangements | Java | Basic Math + Intuitive | java-basic-math-intuitive-by-dz_adman-83wf | Approach:\nJust use basic math for AreaOfRectangle = height * width\nNow all we need is height and width for each cell while calculating area (that cell being t | dz_adman | NORMAL | 2022-06-27T11:10:47.002769+00:00 | 2022-06-27T11:10:47.002832+00:00 | 169 | false | **Approach**:\nJust use basic math for **AreaOfRectangle = height * width**\nNow all we need is **height and width for each cell while calculating area** (that cell being the bottom right end of the rectangle being considered)\n\nThis can be done in following 2 major steps:\n1. Modify matrix values to keep track of max... | 2 | 0 | ['Math'] | 0 |
largest-submatrix-with-rearrangements | Python O(MN log N) with comments | python-omn-log-n-with-comments-by-vsavki-ais3 | \nfrom collections import Counter\n\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n M = len(matrix)\n N = len( | vsavkin | NORMAL | 2021-06-06T11:37:43.750437+00:00 | 2021-06-06T11:37:43.750483+00:00 | 290 | false | ```\nfrom collections import Counter\n\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n M = len(matrix)\n N = len(matrix[0])\n \n colcons = [] # preprocess columns\n for c in range(N):\n cons = []\n s = 0\n for r in r... | 2 | 0 | ['Sorting', 'Python'] | 0 |
largest-submatrix-with-rearrangements | SIMPLE SOLUTION 6MS FASTER THAN 100% | simple-solution-6ms-faster-than-100-by-s-n4us | JAVA CODE IS:\n# \n\nclass Solution {\n int find(int arr[]){\n Arrays.sort(arr);\n int res=0;\n for(int i=0;i<arr.length;i++)\n | shivam_gupta_ | NORMAL | 2021-03-18T06:51:03.900907+00:00 | 2021-03-18T06:51:03.900938+00:00 | 143 | false | JAVA CODE IS:\n# \n```\nclass Solution {\n int find(int arr[]){\n Arrays.sort(arr);\n int res=0;\n for(int i=0;i<arr.length;i++)\n res=Math.max(res,arr[i]*(arr.length-i));\n return res;\n }\n public int largestSubmatrix(int[][] matrix) {\n //size of matrix is n*... | 2 | 0 | [] | 1 |
largest-submatrix-with-rearrangements | [Python] prefix sum, sort rows, compute areas - O(m*nlogn) runtime - with explanation | python-prefix-sum-sort-rows-compute-area-ugi4 | \nApproach: O(m\*nlogn) runtime O(1) space - 1116ms (97.79%), 40.7MB (47.17%)\nCompute prefix from top to bottom. Reset everytime we reach a 0.\nNow go to each | vikktour | NORMAL | 2021-01-23T17:41:24.036855+00:00 | 2024-09-27T23:03:53.442573+00:00 | 444 | false | \nApproach: O(m\\*nlogn) runtime O(1) space - 1116ms (97.79%), 40.7MB (47.17%)\nCompute prefix from top to bottom. Reset everytime we reach a 0.\nNow go to each row and compute area for each value (starting from right to left). For each value, we will use all values to the right of it (including itself) to compute area... | 2 | 0 | ['Sorting', 'Prefix Sum', 'Python', 'Python3'] | 1 |
largest-submatrix-with-rearrangements | Java Solution | O(m * n * logn) time complexity | java-solution-om-n-logn-time-complexity-glk55 | \nclass Solution {\n public int largestSubmatrix(int[][] matrix) {\n \n int m = matrix.length;\n int n = matrix[0].length;\n \n | vishal_sherawat | NORMAL | 2021-01-21T08:08:08.959956+00:00 | 2023-11-26T07:02:40.747236+00:00 | 337 | false | ```\nclass Solution {\n public int largestSubmatrix(int[][] matrix) {\n \n int m = matrix.length;\n int n = matrix[0].length;\n \n for(int i = 1; i < m; i++){\n \n for(int j = 0; j < n; j++){\n \n if(matrix[i][j] == 1){\n ... | 2 | 0 | ['Java'] | 0 |
largest-submatrix-with-rearrangements | O(n^2log(n)) 100% better time complexity[C++} | on2logn-100-better-time-complexityc-by-a-tcy3 | Detail Description Is In Code Itself\nIf You like My Code Then Please Upvot\n\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) | anil111 | NORMAL | 2021-01-18T04:46:23.777348+00:00 | 2021-01-18T04:54:08.598538+00:00 | 81 | false | **Detail Description Is In Code Itself**\nIf You like My Code Then Please Upvot\n```\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) \n {\n int row=matrix.size(),col=matrix[0].size();\n int sum[row][col];\n // calculate no of 1\'s continue in the column\n ... | 2 | 0 | [] | 0 |
largest-submatrix-with-rearrangements | [Python] Clear Solution with Video Explanation | python-clear-solution-with-video-explana-wu53 | Video with clear visualization and explanation:\nhttps://youtu.be/-YVsiCrbeRE\n\n\n\nIntuition: Largest rectangle in histogram with reversed sort\n\n\nCode\n\nc | iverson52000 | NORMAL | 2021-01-17T22:30:15.723926+00:00 | 2021-01-17T22:30:15.723955+00:00 | 237 | false | Video with clear visualization and explanation:\nhttps://youtu.be/-YVsiCrbeRE\n\n\n\nIntuition: Largest rectangle in histogram with reversed sort\n\n\n**Code**\n```\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n n_r, n_c, res = len(matrix), len(matrix[0]), 0\n\n for r ... | 2 | 0 | ['Python'] | 0 |
largest-submatrix-with-rearrangements | C++|| Implementation || DP used | c-implementation-dp-used-by-wargraver-cj0a | \nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) {\n int i,j,k,n,m,ct=0,ans=0;\n n=matrix.size();m=matrix[0].siz | wargraver | NORMAL | 2021-01-17T05:34:02.737627+00:00 | 2021-01-17T05:34:02.737683+00:00 | 66 | false | ```\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& matrix) {\n int i,j,k,n,m,ct=0,ans=0;\n n=matrix.size();m=matrix[0].size();\n int dp[n+5][m+5];\n for(j=0;j<m;j++){\n int flag=0,ct=0;\n for(i=n-1;i>=0;i--){\n if(matrix[i][j]==... | 2 | 0 | [] | 0 |
largest-submatrix-with-rearrangements | SIMPLE SORT, beats 100% easy sorting solution | simple-sort-beats-100-easy-sorting-solut-qabw | `\nclass Solution {\npublic:\n int maxArea(vector<vector<int>> mat,int R,int C) \n{ \n \n vector<vector<int>> hist(R,vector<int> (C,0));\n \n for (i | rdas18246 | NORMAL | 2021-01-17T04:32:07.999880+00:00 | 2021-01-17T04:32:07.999917+00:00 | 61 | false | ````\nclass Solution {\npublic:\n int maxArea(vector<vector<int>> mat,int R,int C) \n{ \n \n vector<vector<int>> hist(R,vector<int> (C,0));\n \n for (int i = 0; i < C; i++) { \n \n hist[0][i] = mat[0][i]; \n \n \n for (int j = 1; j < R; j++) \n hist[j][i] = (mat[j][i] == 0) ? ... | 2 | 0 | [] | 0 |
largest-submatrix-with-rearrangements | [Python] scanning row strategy | python-scanning-row-strategy-by-21eleven-6kb8 | Make a scanning row (r) that you will compare with each row in the matrix. This scanning row tracks the "streak" for each column (how many previous rows had the | 21eleven | NORMAL | 2021-01-17T04:26:54.849160+00:00 | 2021-01-17T15:50:30.484596+00:00 | 148 | false | Make a scanning row (`r`) that you will compare with each row in the matrix. This scanning row tracks the "streak" for each column (how many previous rows had the bit set in that column). If a row does has a zero for a column the scanning row at that column position gets reset to zero, otherwise the streak count at tha... | 2 | 0 | [] | 1 |
largest-submatrix-with-rearrangements | Vertical Profiling with const input | vertical-profiling-with-const-input-by-m-unhm | Still a O(RC) but will less operations than required if modifying the matrix.Code | michelusa | NORMAL | 2025-03-21T00:09:21.042536+00:00 | 2025-03-21T00:09:21.042536+00:00 | 4 | false |
Still a O(RC) but will less operations than required if modifying the matrix.
# Code
```cpp []
class Solution {
public:
int largestSubmatrix(const std::vector<std::vector<int>>& matrix) {
const size_t R = matrix.size();
const size_t C = matrix[0].size();
int max_area = 0;
std::v... | 1 | 0 | ['C++'] | 0 |
largest-submatrix-with-rearrangements | Best Solution | C++ | best-solution-c-by-nssvaishnavi-pyq1 | IntuitionTo solve this problem optimally, we need to make the best use of the column rearrangement to maximize the area of a submatrix consisting entirely of 1s | nssvaishnavi | NORMAL | 2025-02-06T18:10:23.611413+00:00 | 2025-02-06T18:10:23.611413+00:00 | 21 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
To solve this problem optimally, we need to make the best use of the column rearrangement to maximize the area of a submatrix consisting entirely of 1s
# Approach
<!-- Describe your approach to solving the problem. -->
We can treat each ro... | 1 | 0 | ['C++'] | 0 |
largest-submatrix-with-rearrangements | In-place solutions, Beats >90% | in-place-solutions-beats-90-by-simolekc-sszb | Code | simolekc | NORMAL | 2025-01-28T12:28:26.513215+00:00 | 2025-01-28T12:28:26.513215+00:00 | 6 | false |
# Code
```javascript []
/**
* @param {number[][]} matrix
* @return {number}
*/
var largestSubmatrix = function (matrix) {
let n = matrix.length;
let m = matrix[0].length;
for (let i = 1; i < n; i++) {
for (let j = 0; j < m; j++) {
if (matrix[i][j]) matrix[i][j] += matrix[i - 1][j];... | 1 | 0 | ['JavaScript'] | 0 |
largest-submatrix-with-rearrangements | Brute force solution | brute-force-solution-by-execrox-08cu | Code\n\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n m, n, ans = len(matrix), len(matrix[0]), 0\n \n | execrox | NORMAL | 2024-04-26T19:10:16.666172+00:00 | 2024-04-26T19:10:57.843303+00:00 | 4 | false | # Code\n```\nclass Solution:\n def largestSubmatrix(self, matrix: List[List[int]]) -> int:\n m, n, ans = len(matrix), len(matrix[0]), 0\n \n for j in range(n):\n for i in range(1, m):\n matrix[i][j] += matrix[i-1][j] if matrix[i][j] else 0\n \n for... | 1 | 0 | ['Matrix', 'Python3'] | 1 |
largest-submatrix-with-rearrangements | Easyest way to Largest Submatrix With Rearrangements. | easyest-way-to-largest-submatrix-with-re-7bkb | 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 | saslo0165 | NORMAL | 2023-11-27T13:42:13.188207+00:00 | 2023-11-27T13:42:13.188243+00:00 | 21 | 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 | ['Dart'] | 0 |
largest-submatrix-with-rearrangements | [JS] DP (w/Explanation) | Beats 90.91% & 72.73% | js-dp-wexplanation-beats-9091-7273-by-ja-hxuh | Intuition\nUpon encountering this problem, the first idea is to maximize the area of 1s in a binary matrix by rearranging the columns. A key realization is that | jamesBaxtuh | NORMAL | 2023-11-26T22:30:30.563077+00:00 | 2023-11-28T00:59:25.049087+00:00 | 5 | false | # Intuition\nUpon encountering this problem, the first idea is to maximize the area of 1s in a binary matrix by rearranging the columns. A key realization is that we can count the number of 1s in each column and then sort these counts row by row. This sorting effectively simulates rearranging the columns so that rows w... | 1 | 0 | ['Dynamic Programming', 'Greedy', 'Sorting', 'Matrix', 'JavaScript'] | 0 |
largest-submatrix-with-rearrangements | Easy Solution - Ashutosh Kumar | easy-solution-ashutosh-kumar-by-ak180920-zq2v | Intuition\nThe basic intuition to solve this question is first see that we can only swap columns. This means the vertical order of the 1s doesn\'t change. \n\n# | ak18092000 | NORMAL | 2023-11-26T20:57:44.076189+00:00 | 2023-11-26T20:57:44.076230+00:00 | 45 | false | # Intuition\nThe basic intuition to solve this question is first see that we can only swap columns. This means the vertical order of the 1s doesn\'t change. \n\n# Approach\nWe calculate the no. of consecutive 1s vertically in each row and update the elements according to that. The count resets to 0 whenever we encounte... | 1 | 0 | ['C++'] | 0 |
largest-submatrix-with-rearrangements | Sorting + expand consecutive 1s || Faster than 96% || O(n*m*m) || Clean Java Code | sorting-expand-consecutive-1s-faster-tha-hgds | Complexity\n- Time complexity: O(nmm)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n\n# | youssef1998 | NORMAL | 2023-11-26T18:38:58.045329+00:00 | 2023-11-26T18:38:58.045353+00:00 | 132 | false | # Complexity\n- Time complexity: $$O(n*m*m)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int largestSubmatrix(int[][] matrix) {\n int maxArea = 0, n = matrix.length, m ... | 1 | 0 | ['Sorting', 'Java'] | 0 |
largest-submatrix-with-rearrangements | Beats 100.00% of users with Python in runtime and memory | beats-10000-of-users-with-python-in-runt-hfmp | Approach\nUsing dynamic programming to calculate the height of each column, representing the number of consecutive ones ending at that column.\nThen sorting the | anushkaTonk | NORMAL | 2023-11-26T17:58:21.400679+00:00 | 2023-11-26T17:58:21.400701+00:00 | 17 | false | # Approach\nUsing dynamic programming to calculate the height of each column, representing the number of consecutive ones ending at that column.\nThen sorting the columns based on their height.\nThen, calculating the maximum area of the submatrix containing only ones.\n\n# Complexity\n- Time complexity:\nO(m * n * log(... | 1 | 0 | ['Python'] | 0 |
largest-submatrix-with-rearrangements | Unique solution || bests 100% || Well Expalined and easy to understand | unique-solution-bests-100-well-expalined-pm6s | \n# Approach\n\n\n1. Iterate through the matrix, updating each element to represent the height of consecutive 1s above it.\n2. Sort the heights of each row in a | DeepakVrma | NORMAL | 2023-11-26T17:46:53.528418+00:00 | 2023-11-26T17:46:53.528447+00:00 | 160 | false | \n# Approach\n\n\n1. Iterate through the matrix, updating each element to represent the height of consecutive 1s above it.\n2. Sort the heights of each row in ascending order.\n3. Iterate through each row, calculating the area for each height and width combination, updating the maximum area encountered.\n4. Return the ... | 1 | 0 | ['Sorting', 'C++'] | 2 |
largest-submatrix-with-rearrangements | Java Easiest Solution | java-easiest-solution-by-shivansu_7-mstk | 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 | Shivansu_7 | NORMAL | 2023-11-26T17:30:40.895917+00:00 | 2023-11-26T17:30:40.895948+00:00 | 10 | 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 | ['Java'] | 0 |
largest-submatrix-with-rearrangements | Java 100% Efficient code || comment line by line understable | java-100-efficient-code-comment-line-by-d3aux | \n\n# Code\n```\nclass Solution\n {\n \n // Function to find the size of the largest submatrix containing only 1s\n \n public int largestSubmatrix(int[][] m | Dheerajkumar9631r | NORMAL | 2023-11-26T16:36:24.904666+00:00 | 2023-11-26T16:37:41.659803+00:00 | 12 | false | \n\n# Code\n```\nclass Solution\n {\n \n // Function to find the size of the largest submatrix containing only 1s\n \n public int largestSubmatrix(int[][] matrix) {\n // Get the number of rows and columns in the matrix\n \n int m = matrix.length;\n int n = matrix[0].length;\n \n //... | 1 | 0 | ['Java'] | 0 |
largest-submatrix-with-rearrangements | Easy Python Solution - Beats 100% | easy-python-solution-beats-100-by-u_day-aclb | Intuition\nThe problem requires finding the largest submatrix within the given matrix, where every element of the submatrix is 1, after reordering the columns o | U_day | NORMAL | 2023-11-26T16:35:16.210907+00:00 | 2023-11-26T16:35:16.210932+00:00 | 22 | false | # Intuition\nThe problem requires finding the largest submatrix within the given matrix, where every element of the submatrix is 1, after reordering the columns optimally.\n# Approach\n- Initialize variables m and n to store the number of rows and columns in the matrix, respectively.\n- Iterate over the matrix from the... | 1 | 0 | ['Python'] | 0 |
largest-submatrix-with-rearrangements | Dynamic programming approach | dynamic-programming-approach-by-tanmaybh-60b3 | \n# Approach\n Describe your approach to solving the problem. \nThe current approach iterates over each row and column, updating the matrix values and then sort | tanmaybhujade | NORMAL | 2023-11-26T16:08:43.137775+00:00 | 2023-11-26T16:08:43.137803+00:00 | 7 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe current approach iterates over each row and column, updating the matrix values and then sorting for each row. To make it more efficient, we can keep track of the height of consecutive ones for each cell in a separate array. This eliminates the n... | 1 | 0 | ['Java'] | 0 |
largest-submatrix-with-rearrangements | 213ms Beats 100.00%of users with C | 213ms-beats-10000of-users-with-c-by-salm-05zi | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nHistogram Calculation:\ | SalmaHisham | NORMAL | 2023-11-26T14:26:53.376547+00:00 | 2023-11-26T14:26:53.376569+00:00 | 14 | 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**Histogram Calculation:**\nFor each element in the matrix, the algorithm computes the height of the column up to that element by accumulating the consecutive occurren... | 1 | 0 | ['C'] | 0 |
largest-submatrix-with-rearrangements | Runtime 100% | Intuitive aproach | runtime-100-intuitive-aproach-by-user876-70no | Approach\n Describe your approach to solving the problem. \nFor each cell in column we count how long streak of 1 is above them.\n\nThen we are checking row by | user8764IR | NORMAL | 2023-11-26T13:01:33.486547+00:00 | 2023-11-26T13:01:33.486572+00:00 | 86 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nFor each cell in column we count how long streak of `1` is above them.\n\nThen we are checking row by row, how we can replace columns so they will form the best submatrix of `1` -> we can acomplish it by sorting the row. \nNow we need to check which c... | 1 | 0 | ['Java'] | 0 |
largest-submatrix-with-rearrangements | Ruby || O(m*n*log(n))/O(1) | ruby-omnlogno1-by-alecn2002-k3az | \n# Complexity\n- Time complexity:\nO(mnlog(n))\n\n- Space complexity:\nO(1)\n\n# Code\nruby\nclass Numeric\n def max(v) = (self < v ? v : self)\nend\n\ndef | alecn2002 | NORMAL | 2023-11-26T12:00:35.236982+00:00 | 2023-11-26T12:00:35.237015+00:00 | 7 | false | \n# Complexity\n- Time complexity:\n$$O(m*n*log(n))$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```ruby\nclass Numeric\n def max(v) = (self < v ? v : self)\nend\n\ndef largest_submatrix(matrix)\n h, w = matrix.size, matrix.first.size\n matrix.each_cons(2).inject(matrix.first.filter(&:positive?).size) {|res, ... | 1 | 0 | ['Ruby'] | 0 |
largest-submatrix-with-rearrangements | accumulate the height | accumulate-the-height-by-mr_stark-uvd6 | \nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& mt) {\n \n int n = mt.size();\n int m = mt[0].size();\n int r | mr_stark | NORMAL | 2023-11-26T11:52:07.968869+00:00 | 2023-11-26T11:52:38.401563+00:00 | 23 | false | ```\nclass Solution {\npublic:\n int largestSubmatrix(vector<vector<int>>& mt) {\n \n int n = mt.size();\n int m = mt[0].size();\n int res = 0;\n for(int i = 0;i<n;i++)\n {\n for(int j=0;j<m;j++)\n {\n if(mt[i][j] == 1 && i>0)\n ... | 1 | 0 | ['C'] | 0 |
largest-submatrix-with-rearrangements | Swift solution | swift-solution-by-azm819-jufu | Complexity\n- Time complexity: O(m * n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n)\n Add your space complexity here, e.g. O(n) \n\n | azm819 | NORMAL | 2023-11-26T11:47:03.219979+00:00 | 2023-11-26T11:47:03.220010+00:00 | 16 | false | # Complexity\n- Time complexity: $$O(m * n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n func largestSubmatrix(_ matrix: [[Int]]) -> Int {\n var result = 0\n\n var prev... | 1 | 0 | ['Array', 'Greedy', 'Swift', 'Sorting', 'Matrix'] | 0 |
largest-submatrix-with-rearrangements | Java Solution for Largest Submatrix With Rearrangements Problem | java-solution-for-largest-submatrix-with-co1q | Intuition\n Describe your first thoughts on how to solve this problem. \nThe approach aims to find the largest submatrix containing only 1s after rearranging th | Aman_Raj_Sinha | NORMAL | 2023-11-26T11:28:24.118896+00:00 | 2023-11-26T11:28:24.118923+00:00 | 106 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe approach aims to find the largest submatrix containing only 1s after rearranging the columns optimally. It works by first computing the height of each column by counting the consecutive 1s from the top. Then, it sorts each row\u2019s ... | 1 | 0 | ['Java'] | 0 |
largest-submatrix-with-rearrangements | Typescript: sort and without sort solutions (Runtime beats 100%) | typescript-sort-and-without-sort-solutio-jhcg | Complexity\n- Time complexity:\nO(n * m)\n\n- Space complexity:\nO(n)\n\n# Code\n\n// TC (m * n), SC: (n)\nfunction largestSubmatrix(matrix: number[][]): number | alexgavrilov | NORMAL | 2023-11-26T09:57:21.787438+00:00 | 2023-11-26T09:57:21.787464+00:00 | 50 | false | # Complexity\n- Time complexity:\nO(n * m)\n\n- Space complexity:\nO(n)\n\n# Code\n```\n// TC (m * n), SC: (n)\nfunction largestSubmatrix(matrix: number[][]): number {\n const m: number = matrix.length;\n const n: number = matrix[0].length;\n let maxArea: number = 0;\n let prevHeights: [number, number][] = ... | 1 | 0 | ['Array', 'Greedy', 'Sorting', 'Matrix', 'TypeScript', 'JavaScript'] | 0 |
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