Set Matrix Zeroes medium

Problem Statement

Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0. You must do it in place.

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]]

Steps and Explanation

The naive approach would be to iterate through the matrix, and whenever we find a 0, iterate again to set all elements in that row and column to 0. This has a time complexity of O(mn (m+n)). To improve this, we'll use the first row and first column as markers.

1. Marker Rows and Columns: We use the first row and the first column to store information about whether a row or column needs to be zeroed. If we find a 0 at matrix[i][j], we set matrix[i][0] and matrix[0][j] to 0. This acts as a flag.

2. Iterate: We iterate through the matrix from matrix[1][1] to matrix[m-1][n-1]. If matrix[i][j] is 0, then we set matrix[i][0] and matrix[0][j] to 0.

3. Zero Rows and Columns: After the iteration, we use the first row and column as markers. If matrix[i][0] is 0, we set the entire i-th row to 0. Similarly, if matrix[0][j] is 0, we set the entire j-th column to 0.

4. Handle First Row and Column: Finally, we handle the first row and column separately, based on whether matrix[0][0] is 0.

This approach reduces the time complexity significantly because we only iterate through the matrix a constant number of times.

Code (Java)

class Solution {
    public void setZeroes(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        boolean firstRowZero = false;
        boolean firstColZero = false;

        // Check if the first row and column need to be zeroed
        for (int i = 0; i < m; i++) {
            if (matrix[i][0] == 0) {
                firstColZero = true;
                break;
            }
        }
        for (int j = 0; j < n; j++) {
            if (matrix[0][j] == 0) {
                firstRowZero = true;
                break;
            }
        }

        // Mark rows and columns to be zeroed
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[i][j] == 0) {
                    matrix[i][0] = 0;
                    matrix[0][j] = 0;
                }
            }
        }

        // Zero rows and columns based on markers
        for (int i = 1; i < m; i++) {
            if (matrix[i][0] == 0) {
                for (int j = 0; j < n; j++) {
                    matrix[i][j] = 0;
                }
            }
        }
        for (int j = 1; j < n; j++) {
            if (matrix[0][j] == 0) {
                for (int i = 0; i < m; i++) {
                    matrix[i][j] = 0;
                }
            }
        }

        //Handle first row and column
        if(firstRowZero){
            for(int j=0; j<n; j++) matrix[0][j] = 0;
        }
        if(firstColZero){
            for(int i=0; i<m; i++) matrix[i][0] = 0;
        }
    }
}

Complexity

• Time Complexity: O(m*n) - We iterate through the matrix twice.
• Space Complexity: O(1) - We use constant extra space. We modify the matrix in place.