This approach uses two additional arrays to track the rows and columns that need to be zeroed.
- Traverse the matrix and mark rows and columns that need to be zeroed.
- Then, update the matrix based on the values in these arrays.
Time Complexity: O(m * n), where m is the number of rows and n is the number of columns.
Space Complexity: O(m + n) for the additional arrays.
1#include <stdio.h>
2#include <stdbool.h>
3
4void setZeroes(int** matrix, int matrixSize, int* matrixColSize){
5 bool row[matrixSize];
6 bool col[*matrixColSize];
7
8 for(int i = 0; i < matrixSize; ++i) row[i] = false;
9 for(int i = 0; i < *matrixColSize; ++i) col[i] = false;
10
11 for (int i = 0; i < matrixSize; ++i) {
12 for (int j = 0; j < *matrixColSize; ++j) {
13 if (matrix[i][j] == 0) {
14 row[i] = true;
15 col[j] = true;
16 }
17 }
18 }
19
20 for (int i = 0; i < matrixSize; ++i) {
21 for (int j = 0; j < *matrixColSize; ++j) {
22 if (row[i] || col[j]) {
23 matrix[i][j] = 0;
24 }
25 }
26 }
27}This approach uses the matrix itself to track zeroes.
- Use the first row and column as markers.
- Traverse the matrix and if an element is zero, mark its row and column.
- Traverse again and set matrix elements to zero based on marked rows and columns.
Time Complexity: O(m * n), where m is the number of rows and n is the number of columns.
Space Complexity: O(1) since the algorithm runs in constant space.
1public class Solution {
2 public void SetZeroes(int[][] matrix) {
3 int m = matrix.Length;
4 int n = matrix[0].Length;
5 bool isCol = false;
6 for (int i = 0; i < m; i++) {
7 if (matrix[i][0] == 0) isCol = true;
8 for (int j = 1; j < n; j++) {
9 if (matrix[i][j] == 0) {
10 matrix[i][0] = 0;
11 matrix[0][j] = 0;
12 }
13 }
14 }
15
16 for (int i = 1; i < m; i++) {
17 for (int j = 1; j < n; j++) {
18 if (matrix[i][0] == 0 || matrix[0][j] == 0) {
19 matrix[i][j] = 0;
20 }
21 }
22 }
23
24 if (matrix[0][0] == 0) {
25 for (int j = 0; j < n; j++) {
26 matrix[0][j] = 0;
27 }
28 }
29
30 if (isCol) {
31 for (int i = 0; i < m; i++) {
32 matrix[i][0] = 0;
33 }
34 }
35 }
36}