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# 73. Set Matrix Zeroes

### Description

 Given a *m* x *n* matrix, if an element is 0, set its entire row and column to 0. Do it [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm).

### **Constraints**

### **Approach**

### Links

* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/set-matrix-zeroes)
* ProgramCreek

### Examples

{% tabs %}
{% tab title="Example 1" %}
**Input:** \[ \[1, 1, 1], \[1, 0, 1], \[1, 1, 1] ]

**Output:** \[ \[1, 0, 1], \[0, 0, 0], \[1, 0, 1] ]
{% endtab %}

{% tab title="Example 2" %}
**Input:** \[ \[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] ]
{% endtab %}
{% endtabs %}

### Solutions

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O((M×N)×(M+N)) where M and N are the number of rows and columns 
 *    respectively. Even though this solution avoids using space, but is very 
 *    inefficient since in worst case for every cell we might have to zero out its 
 *    corresponding row and column. Thus for all (M×N) cells zeroing out (M+N) cells.
 * Space complexity : O(1)
 */

class Solution {
    public void setZeroes(int[][] matrix) {
        int row = matrix.length;
        int col = matrix[0].length;
        int MAX = -1000000;
        for(int i = 0; i < row; i++) {
            for(int j = 0; j < col; j++) {
                if(matrix[i][j] == 0) {
                    for(int k = 0; k < row; k++) {
                        if(matrix[k][j] != 0) {
                            matrix[k][j] = MAX;
                        }
                    }
                    for(int k = 0; k < col; k++) {
                        if(matrix[i][k] != 0) {
                            matrix[i][k] = MAX;
                        }
                    }
                }
            }
        }
        for(int i = 0; i < row; i++) {
            for(int j = 0; j < col; j++) {
                if(matrix[i][j] == MAX) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : O(M×N)
 * Space complexity : O(1)
 */

class Solution {
    public void setZeroes(int[][] matrix) {
        int row = matrix.length;
        int col = matrix[0].length;
        int col0 = 1;
        for(int i = 0; i < row; i++) {
            if(matrix[i][0] == 0) col0 = 0;
            for(int j = 1; j < col; j++) {
                if(matrix[i][j] == 0) {
                    matrix[i][0] = 0;
                    matrix[0][j] = 0;
                }
            }
        }
        for(int i = row-1; i >= 0; i--) {
            for(int j = col-1; j >= 1; j--) {
                if(matrix[i][0] == 0 || matrix[0][j] == 0) {
                    matrix[i][j] = 0;
                }
            }
            if(col0 == 0) matrix[i][0] = 0;
        }
    }
}
```

{% endtab %}
{% endtabs %}

### Follow up

* A straight forward 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?


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