Cakeminator | Editorial by khushalagarwal

In this problem we need to find the row or the column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before.Thus first we create two sets
each for row and column and insert in it the row and column in which we encounter the letter 'S' .
The we do a binary search for the row and column correspondingly which is not in the set and thus put at that location any letter/character (here '1') of our choice so that we can count the number.
finally we iterate through the whole array and count the number of location which has the character(here '1')

Here the constraints are small so this code can be easily solve through this approach.

```
#include<bits/stdc++.h>
using namespace std;
int main() {
    int m, n;
    cin >> m >> n;
    char arr[m][n];
    set<int> row, col;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            cin >> arr[i][j];
        }
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (arr[i][j] == 'S') {
                row.insert(i);
                col.insert(j);
            }
        }
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (!binary_search(row.begin(), row.end(), i)) {
                arr[i][j] = '1';
            }
        }
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (!binary_search(col.begin(), col.end(), j)) {
                arr[i][j] = '1';
            }
        }
    }
    int ctr = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (arr[i][j] == '1')
                ctr++;
        }
    }
    cout << ctr;
}
```

Cakeminator | Editorial by khushalagarwal

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