Fredo and Game | Editorial by aksingh007

The problem statement itself provides the solution too. We basically just have to implement whatever is written in the problem.
We are given an array  **Arr[]** and an integer** A**. Each element  of  will be either 0 or 1. We traverse from the start of the array till the end. At any moment, if we
encounter , **Arr[i]=1** we increase the value of A by 2 else if **Arr[i]=0** , we decrease the value of A by 1.As soon as **A=0**, , we stop traversing further. If we are
able to reach the end, we output the value of A, else we output the index of the array where we stopped.

**Code:** Java

```
import java.util.*;
class TestClass {
    public static void main(String args[] ) throws Exception {
       Scanner sc=new Scanner(System.in);
       int t=sc.nextInt();
       for(int p=0;p<t;p++){
           int amino=sc.nextInt();
           int n=sc.nextInt();
           int[] arr=new int[100005];
           int i;
           for( i=1;i<=n;i++)
             arr[i]=sc.nextInt();
           for( i=1;i<=n;i++){
               if(arr[i]==1)
               amino+=2;
               else
               amino--;
               if(amino==0)
                   break;
           }
           if(amino==0 && i<n)
              System.out.println("No "+i);
              else
              //if(i==n+1)
                System.out.println("Yes "+amino);
       } 
    }
}
```

Fredo and Game | Editorial by aksingh007

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