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.
1import java.util.*;
2
3public class MatrixZero {
4 public void setZeroes(int[][] matrix) {
5 int m = matrix.length;
6 int n = matrix[0].length;
7 boolean[] row = new boolean[m];
8 boolean[] col = new boolean[n];
9
10 for (int i = 0; i < m; i++) {
11 for (int j = 0; j < n; j++) {
12 if (matrix[i][j] == 0) {
13 row[i] = true;
14 col[j] = true;
15 }
16 }
17 }
18
19 for (int i = 0; i < m; i++) {
20 for (int j = 0; j < n; j++) {
21 if (row[i] || col[j]) {
22 matrix[i][j] = 0;
23 }
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.
1def setZeroes(matrix):
2 m, n = len(matrix), len(matrix[0])
3 col0 = 1
4 for i in range(m):
5 if matrix[i][0] == 0:
6 col0 = 0
7 for j in range(1, n):
8 if matrix[i][j] == 0:
9 matrix[i][0] = 0
10 matrix[0][j] = 0
11
12 for i in range(1, m):
13 for j in range(1, n):
14 if matrix[i][0] == 0 or matrix[0][j] == 0:
15 matrix[i][j] = 0
16
17 if matrix[0][0] == 0:
18 for j in range(n):
19 matrix[0][j] = 0
20
21 if col0 == 0:
22 for i in range(m):
23 matrix[i][0] = 0