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package g0001_0100.s0073_set_matrix_zeroes;

// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Hash_Table #Matrix
// #Udemy_2D_Arrays/Matrix #Big_O_Time_O(m*n)_Space_O(1)
// #2023_08_11_Time_1_ms_(79.07%)_Space_44.4_MB_(94.19%)

/**
 * 73 - Set Matrix Zeroes\.
 *
 * Medium
 *
 * Given an `m x n` integer matrix `matrix`, if an element is `0`, set its entire row and column to `0`'s, and return _the matrix_.
 *
 * You must do it [in place](https://en.wikipedia.org/wiki/In-place_algorithm).
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/08/17/mat1.jpg)
 *
 * **Input:** matrix = \[\[1,1,1],[1,0,1],[1,1,1]]
 *
 * **Output:** [[1,0,1],[0,0,0],[1,0,1]] 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/08/17/mat2.jpg)
 *
 * **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
 *
 * **Follow up:**
 *
 * *   A straightforward 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?
**/
public class Solution {
    // Approach: Use first row and first column for storing whether in future
    //           the entire row or column needs to be marked 0
    public void setZeroes(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        boolean row0 = false;
        boolean col0 = false;
        // Check if 0th col needs to be market all 0s in future
        for (int[] ints : matrix) {
            if (ints[0] == 0) {
                col0 = true;
                break;
            }
        }
        // Check if 0th row needs to be market all 0s in future
        for (int i = 0; i < n; i++) {
            if (matrix[0][i] == 0) {
                row0 = true;
                break;
            }
        }
        // Store the signals in 0th row and column
        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;
                }
            }
        }
        // Mark 0 for all cells based on signal from 0th row and 0th column
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[i][0] == 0 || matrix[0][j] == 0) {
                    matrix[i][j] = 0;
                }
            }
        }
        // Set 0th column
        for (int i = 0; i < m; i++) {
            if (col0) {
                matrix[i][0] = 0;
            }
        }
        // Set 0th row
        for (int i = 0; i < n; i++) {
            if (row0) {
                matrix[0][i] = 0;
            }
        }
    }
}