73. Set Matrix Zeroes

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.

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

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?

Solutions

Solution 1: Array Mark

We use arrays rows and cols to mark the rows and columns to be cleared.

Then traverse the matrix again, and clear the elements in the rows and columns marked in rows and cols.

The time complexity is $O(m\times n)$, and the space complexity is $O(m+n)$. Where $m$ and $n$ are the number of rows and columns of the matrix respectively.

Python Code

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class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        m, n = len(matrix), len(matrix[0])
        rows = [0] * m
        cols = [0] * n
        for i, row in enumerate(matrix):
            for j, v in enumerate(row):
                if v == 0:
                    rows[i] = cols[j] = 1
        for i in range(m):
            for j in range(n):
                if rows[i] or cols[j]:
                    matrix[i][j] = 0

Java Code

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class Solution {
    public void setZeroes(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        boolean[] rows = new boolean[m];
        boolean[] cols = new boolean[n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) {
                    rows[i] = true;
                    cols[j] = true;
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (rows[i] || cols[j]) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
}

C++ Code

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class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        vector<bool> rows(m);
        vector<bool> cols(n);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (!matrix[i][j]) {
                    rows[i] = 1;
                    cols[j] = 1;
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (rows[i] || cols[j]) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
};

Go Code

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func setZeroes(matrix [][]int) {
	m, n := len(matrix), len(matrix[0])
	rows := make([]bool, m)
	cols := make([]bool, n)
	for i, row := range matrix {
		for j, v := range row {
			if v == 0 {
				rows[i] = true
				cols[j] = true
			}
		}
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if rows[i] || cols[j] {
				matrix[i][j] = 0
			}
		}
	}
}

TypeScript Code

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/**
 Do not return anything, modify matrix in-place instead.
 */
function setZeroes(matrix: number[][]): void {
    const m = matrix.length;
    const n = matrix[0].length;
    const rows: boolean[] = new Array(m).fill(false);
    const cols: boolean[] = new Array(n).fill(false);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (matrix[i][j] === 0) {
                rows[i] = true;
                cols[j] = true;
            }
        }
    }
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (rows[i] || cols[j]) {
                matrix[i][j] = 0;
            }
        }
    }
}

JavaScript Code

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/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var setZeroes = function (matrix) {
    const m = matrix.length;
    const n = matrix[0].length;
    const rows = new Array(m).fill(false);
    const cols = new Array(n).fill(false);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (matrix[i][j] == 0) {
                rows[i] = true;
                cols[j] = true;
            }
        }
    }
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (rows[i] || cols[j]) {
                matrix[i][j] = 0;
            }
        }
    }
};

C# Code

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public class Solution {
    public void SetZeroes(int[][] matrix) {
        int m = matrix.Length, n = matrix[0].Length;
        bool[] rows = new bool[m], cols = new bool[n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) {
                    rows[i] = true;
                    cols[j] = true;
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (rows[i] || cols[j]) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
}

Solution 2: Mark in Place

In the first method, we use an additional array to mark the rows and columns to be cleared. In fact, we can also use the first row and first column of the matrix to mark them, without creating an additional array.

Since the first row and the first column are used to mark, their values may change due to the mark, so we need additional variables $i0$, $j0$ to mark whether the first row and the first column need to be cleared.

The time complexity is $O(m\times n)$, and the space complexity is $O(1)$. Where $m$ and $n$ are the number of rows and columns of the matrix respectively.