KW6 | Editorial by Uvaish

KW6 | Editorial by Uvaish_Zafri

**Problem:**
>Our University is planning to give DL(Duty Leave) to all the students. I know you got it already but it's happening again.
> But this time they give you a list of N numbers and you need to rearrange the given list and give it back.
>  If a student's list has two adjacent odd numbers or two adjacent even number DL would not be given to him.
>   Since you are a smart student you need to tell whether you will get the DL or not if you arrange the list optimally.

Sample Input:
    **4
    1 1 4 8**
Sample Output:
  **  DL**
Sample Input:
    **4
    1 1 3 4**
Sample Output:
    **DETAIN**
    
    
So, from some simple observations we can see that "DL" is only possible if and only if number of odd or even number are equal to half of N.

so, now let's code the basics in CPP :)

```
#include<bits/stdc++.h>
using namespace std;

int main(){
    int n;
    cin >> n;
    int cnt1=0;
    for(int i=0;i<n;i++){
        int x=0;
        cin >> x;
        cnt1 += (x%2);  // count of odd numbers 
    }
    if(cnt1 == n/2 || (n-cnt1) == n/2){   //checking half condition
        cout<<"DL";
    }
    else{
        cout<<"DETAIN";
    }
    return 0;
}
```

KW6 | Editorial by Uvaish_Zafri

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