The Doors | Editorial by khushalagarwal

The simplest approach for this problem is that firstly count the number of 1's and 0's in the array (let it be ctr1 and ctr2) and finally run a loop through the array and if 1 is encountered ctr1-- and if ctr1 becomes 0 break and display the i+1th position an come out of the loop similarly do it for ctr2.
```
#include <bits/stdc++.h>
using namespace std;
int main() {
    int n;
    cin >> n;
    vector<int> arr;
    for (int i = 0; i < n; i++) {
        int m;
        cin >> m;
        arr.push_back(m);
    }
    int ctr = 0, ctr1 = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] == 0)
            ctr++;
        else
            ctr1++;
    }
    for (int i = 0; i < n; i++) {
        if (arr[i] == 0) {
            ctr--;
            if (ctr == 0) {
                cout << i + 1;
                break;
            }
        }
        if (arr[i] == 1) {
            ctr1--;
            if (ctr1 == 0) {
                cout << i + 1;
                break;
            }
        }
    }
}
```

The Doors | Editorial by khushalagarwal

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