A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end. For example, the matrix diagonal starting from mat[2][0], where mat is a 6 x 3 matrix, includes cells mat[2][0], mat[3][1], and mat[4][2].
Given an m x n matrix mat of integers, sort each matrix diagonal in ascending order and return the resulting matrix.
Example 1:
Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]] Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]
Example 2:
Input: mat = [[11,25,66,1,69,7],[23,55,17,45,15,52],[75,31,36,44,58,8],[22,27,33,25,68,4],[84,28,14,11,5,50]] Output: [[5,17,4,1,52,7],[11,11,25,45,8,69],[14,23,25,44,58,15],[22,27,31,36,50,66],[84,28,75,33,55,68]]
Constraints:
m == mat.lengthn == mat[i].length1 <= m, n <= 1001 <= mat[i][j] <= 100The first approach involves using a hashmap or dictionary to store diagonals. For any element at (i, j), the diagonal can be identified by the key (i - j). This key is unique to all elements on the same diagonal. We then collect each diagonal's elements into a list, sort them, and place them back into the matrix.
This C solution uses a hashmap-like approach by leveraging an array diagonals, where the index represents the diagonal identified by (i - j + 50) to ensure positive index values. We first populate this array with elements from each diagonal, sort the array, and write the sorted values back into the matrix.
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Time Complexity: O(m * n * log(min(m, n))) due to the sorting of up to min(m, n) elements for each diagonal.
Space Complexity: O(m * n) for storing diagonal elements.
This approach aims to perform cross-diagonal sorting directly within the matrix itself, minimizing additional storage usage. By extracting diagonal elements one at a time, sorting them, and reinserting them, we achieve the diagonal sorting task with constrained space complexity.
This C code sorts diagonals in place without extra structures. It iterates over each diagonal from top row and left column, extracts, sorts, and reinserts elements back directly.
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Java
Python
C#
JavaScript
Time Complexity: O(m * n * log(min(m, n))) due to sorting operations.
Space Complexity: O(min(m, n)) for temporary diagonal storage.
| Approach | Complexity |
|---|---|
| Using Hashmap for Diagonals | Time Complexity: |
| In-place Sorting without Extra Storage | Time Complexity: |
Sort the Matrix Diagonally | Live Coding with Explanation | Leetcode #1329 • Algorithms Made Easy • 13,475 views views
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