The 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.
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.
1import java.util.*;
2
3class Solution {
4 public int[][] diagonalSort(int[][] mat) {
5 Map<Integer, PriorityQueue<Integer>> diagonals = new HashMap<>();
6 int m = mat.length, n = mat[0].length;
7
8 for (int i = 0; i < m; i++) {
9 for (int j = 0; j < n; j++) {
10 diagonals.putIfAbsent(i - j, new PriorityQueue<>());
11 diagonals.get(i - j).add(mat[i][j]);
12 }
13 }
14
15 for (int i = 0; i < m; i++) {
16 for (int j = 0; j < n; j++) {
17 mat[i][j] = diagonals.get(i - j).poll();
18 }
19 }
20
21 return mat;
22 }
23}This Java solution utilizes a HashMap to map diagonals to a min heap (PriorityQueue). Elements are added to respective heaps, and then reinserted into the matrix in sorted order by polling from heaps.
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.
Time Complexity: O(m * n * log(min(m, n))) due to sorting operations.
Space Complexity: O(min(m, n)) for temporary diagonal storage.
1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4
5void sort(int *array, int size) {
6 for (int step = 0; step < size - 1; ++step) {
7 for (int i = 0; i < size - step - 1; ++i) {
8 if (array[i] > array[i + 1]) {
9 int temp = array[i];
10 array[i] = array[i + 1];
11 array[i + 1] = temp;
12 }
13 }
14 }
15}
16
17void sortDiagonal(int **mat, int m, int n, int x, int y) {
18 int array[100], idx = 0;
19 int i = x, j = y;
20 while (i < m && j < n) {
21 array[idx++] = mat[i][j];
22 i++; j++;
23 }
24 sort(array, idx);
25 i = x; j = y; idx = 0;
26 while (i < m && j < n) {
27 mat[i][j] = array[idx++];
28 i++; j++;
29 }
30}
31
32int** diagonalSort(int** mat, int matSize, int* matColSize, int* returnSize, int** returnColumnSizes) {
33 *returnSize = matSize;
34 *returnColumnSizes = matColSize;
35
36 for (int i = 0; i < matSize; i++)
37 sortDiagonal(mat, matSize, *matColSize, i, 0);
38
39 for (int j = 1; j < *matColSize; j++)
40 sortDiagonal(mat, matSize, *matColSize, 0, j);
41
42 return mat;
43}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.