leetcode

Set Matrix Zeroes - Link

Question Description

Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0’s.

You must do it in place.

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Constraints

• m == matrix.length
• n == matrix[0].length
• 1 <= m, n <= 200
• -2^31 <= matrix[i][j] <= 2^31 - 1

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Approach

Use the first row and first column as markers to record which rows/columns should be zeroed. Track separately if the first row or first column need to be zeroed to avoid losing marker info. In the second pass, set entire rows/columns to zero based on these markers.

Why this approach works:

• Uses O(1) extra space by utilizing the matrix itself as storage
• First pass marks which rows and columns need to be zeroed
• Second pass actually sets the zeros
• Handles edge cases for first row and first column separately

Alternative approaches considered:

• Could use separate arrays to track rows and columns, but that would use O(m+n) space
• Could use a set to store positions, but that would also use extra space

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Dry Run

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

Step-by-step execution:

• Step 1: Find 0 at position [1][1], mark row 1 and column 1
• Step 2: Set matrix[0][1] = 0 and matrix[1][0] = 0 as markers
• Step 3: Set entire column 1 to 0 (except already processed cells)
• Step 4: Set entire row 1 to 0 (except already processed cells)
• Step 5: Handle first row and first column if needed

Final Answer = [[1,0,1],[0,0,0],[1,0,1]]

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Solution

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

        boolean r0F = false;
        boolean c0F = false;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 && matrix[i][j] == 0) r0F = true;
                if (j == 0 && matrix[i][j] == 0) c0F = true;
                if (matrix[i][j] == 0) {
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
            }
        }

        for (int i = 1; i < n; i++) {
            if (matrix[0][i] == 0) {
                for (int j = 0; j < m; j++) {
                    matrix[j][i] = 0;
                }
            }
        }

        for (int i = 1; i < m; i++) {
            if (matrix[i][0] == 0) {
                for (int j = 0; j < n; j++) {
                    matrix[i][j] = 0;
                }
            }
        }

        if (r0F) {
            for (int i = 0; i < n; i++) {
                matrix[0][i] = 0;
            }
        }

        if (c0F) {
            for (int i = 0; i < m; i++) {
                matrix[i][0] = 0;
            }
        }
    }
}

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Time and Space Complexity

• Time Complexity: O(m * n) - each cell is visited a constant number of times
• Space Complexity: O(1) - in-place, only a few extra variables

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