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.
1function setZeroes(matrix) {
2 let m = matrix.length;
3 let n = matrix[0].length;
4 let row = new Array(m).fill(false);
5 let col = new Array(n).fill(false);
6
7 for (let i = 0; i < m; i++) {
8 for (let j = 0; j < n; j++) {
9 if (matrix[i][j] === 0) {
10 row[i] = true;
11 col[j] = true;
12 }
13 }
14 }
15
16 for (let i = 0; i < m; i++) {
17 for (let j = 0; j < n; j++) {
18 if (row[i] || col[j]) {
19 matrix[i][j] = 0;
20 }
21 }
22 }
23}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