73. Set Matrix Zeroes

Given an m x n matrix. If an element is 0, set its entire row and column to 0. Do it in-place.

Follow up:

• A straight forward solution using O(mn) space is probably a bad idea.

• A simple improvement uses O(m + n) space, but still not the best solution.

• Could you devise a constant space solution?

Example 1:

Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[1,0,1],[0,0,0],[1,0,1]]

Example 2:

Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]

Constraints:

• m == matrix.length

• n == matrix[0].length

• 1 <= m, n <= 200

• -231 <= matrix[i][j] <= 231 - 1

Solution

They are using the first row and column as a memory to keep track of all the 0's in the entire matrix.

But for the first col must be handled (including tack and set) separately.

the inner loop is from j = 1 to n-1. to ignore the first col

the setting loop shall be set from the end to the head, in order to avoid the impack of changing matrix[0][j] & matrix[i][0]

 public void setZeroes(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        int col0 = 1;
        
        for(int i = 0; i < m; i++){
            if(matrix[i][0] == 0) col0 = 0;
            for(int j = 1; j < n; j++){
                if(matrix[i][j] == 0) {
                    matrix[i][0] = 0; matrix[0][j] = 0;
                }
            }
        }
        
        for(int i = m-1; i >= 0; i--){
            for(int j = n-1; j >= 1; j--){
                if(matrix[0][j] == 0 || matrix[i][0] == 0) matrix[i][j] = 0;
                
            }
            if(col0 == 0) matrix[i][0] = 0;
        }
    }