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:
![](https://spcdn.pages.dev/leetcode/problems/0073.Set%20Matrix%20Zeroes/images/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://spcdn.pages.dev/leetcode/problems/0073.Set%20Matrix%20Zeroes/images/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?
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.
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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
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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;
}
}
}
}
}
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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;
}
}
}
}
};
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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
}
}
}
}
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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;
}
}
}
}
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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;
}
}
}
};
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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.
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| class Solution:
def setZeroes(self, matrix: List[List[int]]) -> None:
m, n = len(matrix), len(matrix[0])
i0 = any(v == 0 for v in matrix[0])
j0 = any(matrix[i][0] == 0 for i in range(m))
for i in range(1, m):
for j in range(1, n):
if matrix[i][j] == 0:
matrix[i][0] = matrix[0][j] = 0
for i in range(1, m):
for j in range(1, n):
if matrix[i][0] == 0 or matrix[0][j] == 0:
matrix[i][j] = 0
if i0:
for j in range(n):
matrix[0][j] = 0
if j0:
for i in range(m):
matrix[i][0] = 0
|
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| class Solution {
public void setZeroes(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
boolean i0 = false, j0 = false;
for (int j = 0; j < n; ++j) {
if (matrix[0][j] == 0) {
i0 = true;
break;
}
}
for (int i = 0; i < m; ++i) {
if (matrix[i][0] == 0) {
j0 = true;
break;
}
}
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;
}
}
}
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;
}
}
}
if (i0) {
for (int j = 0; j < n; ++j) {
matrix[0][j] = 0;
}
}
if (j0) {
for (int i = 0; i < m; ++i) {
matrix[i][0] = 0;
}
}
}
}
|
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| class Solution {
public:
void setZeroes(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
bool i0 = false, j0 = false;
for (int j = 0; j < n; ++j) {
if (matrix[0][j] == 0) {
i0 = true;
break;
}
}
for (int i = 0; i < m; ++i) {
if (matrix[i][0] == 0) {
j0 = true;
break;
}
}
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;
}
}
}
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;
}
}
}
if (i0) {
for (int j = 0; j < n; ++j) {
matrix[0][j] = 0;
}
}
if (j0) {
for (int i = 0; i < m; ++i) {
matrix[i][0] = 0;
}
}
}
};
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| func setZeroes(matrix [][]int) {
m, n := len(matrix), len(matrix[0])
i0, j0 := false, false
for j := 0; j < n; j++ {
if matrix[0][j] == 0 {
i0 = true
break
}
}
for i := 0; i < m; i++ {
if matrix[i][0] == 0 {
j0 = true
break
}
}
for i := 1; i < m; i++ {
for j := 1; j < n; j++ {
if matrix[i][j] == 0 {
matrix[i][0], matrix[0][j] = 0, 0
}
}
}
for i := 1; i < m; i++ {
for j := 1; j < n; j++ {
if matrix[i][0] == 0 || matrix[0][j] == 0 {
matrix[i][j] = 0
}
}
}
if i0 {
for j := 0; j < n; j++ {
matrix[0][j] = 0
}
}
if j0 {
for i := 0; i < m; i++ {
matrix[i][0] = 0
}
}
}
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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 i0 = matrix[0].includes(0);
const j0 = matrix.map(row => row[0]).includes(0);
for (let i = 1; i < m; ++i) {
for (let j = 1; j < n; ++j) {
if (matrix[i][j] === 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for (let i = 1; i < m; ++i) {
for (let j = 1; j < n; ++j) {
if (matrix[i][0] === 0 || matrix[0][j] === 0) {
matrix[i][j] = 0;
}
}
}
if (i0) {
matrix[0].fill(0);
}
if (j0) {
for (let i = 0; i < m; ++i) {
matrix[i][0] = 0;
}
}
}
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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;
let i0 = matrix[0].some(v => v == 0);
let j0 = false;
for (let i = 0; i < m; ++i) {
if (matrix[i][0] == 0) {
j0 = true;
break;
}
}
for (let i = 1; i < m; ++i) {
for (let j = 1; j < n; ++j) {
if (matrix[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for (let i = 1; i < m; ++i) {
for (let j = 1; j < n; ++j) {
if (matrix[i][0] == 0 || matrix[0][j] == 0) {
matrix[i][j] = 0;
}
}
}
if (i0) {
for (let j = 0; j < n; ++j) {
matrix[0][j] = 0;
}
}
if (j0) {
for (let i = 0; i < m; ++i) {
matrix[i][0] = 0;
}
}
};
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| public class Solution {
public void SetZeroes(int[][] matrix) {
int m = matrix.Length, n = matrix[0].Length;
bool i0 = matrix[0].Contains(0), j0 = false;
for (int i = 0; i < m; ++i) {
if (matrix[i][0] == 0) {
j0 = true;
break;
}
}
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;
}
}
}
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;
}
}
}
if (i0) {
for (int j = 0; j < n; ++j) {
matrix[0][j] = 0;
}
}
if (j0) {
for (int i = 0; i < m; ++i) {
matrix[i][0] = 0;
}
}
}
}
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