# 73. Set Matrix Zeroes | LeetCode | CodingKaro

## 73. Set Matrix Zeroes

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.

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`

• 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?

The solution to the above question

class Solution {
public:
void setZeroes(vector<vector<int>>& v) {
bool row=false;
bool col=false;
for(int i=0; i<v.size(); i++){
for(int j=0; j<v[0].size(); j++){
if(v[i][j]==0){
v[0][j]=0;
v[i][0]=0;
if(i==0){
row=true;
}
if(j==0){
col=true;
}
}
}
}
for(int i=1; i<v[0].size(); i++){
if(v[0][i]==0){
for(int j=0; j<v.size(); j++){
v[j][i]=0;
}
}
}
for(int i=1; i<v.size(); i++){
if(v[i][0]==0){
for(int j=0; j<v[0].size(); j++){
v[i][j]=0;
}
}
}
if(row){
for(int i=0; i<v[0].size(); i++){
v[0][i]=0;
}
}
if(col){
for(int i=0; i<v.size(); i++){
v[i][0]=0;
}
}

for(int i=0; i<v.size(); i++){
for(int j=0; j<v[0].size(); j++){
{
cout<<v[i][j]<<" ";
}
}
cout<<endl;
}
}
};