Even Array | Editorial by manikasharma

**Approach**
1. We can notice that if the number of even elements at wrong index and number of odd elements at wrong index are same then the
array can be converted to even array otherwise it cannot be converted.
2. Initialize 2 counter variables oddcount=0 and evencount=0
3. Then we traverse the array , If the index **i** is even and the array element present at this index is odd , then increase the counter 
oddcount++
4. If the index **i** is odd and the array element present at this index is even, then increase the counter evencount++
5. If both evencount and oddcount are equal then we know that it is possible to create a even array , Therefore print oddcount 
6. And if the evencount and oddcount are not equal then we can comprehend that creating the even array is impossible , therefore
print -1.

**Code**
```
import java.util.*;
public class Codechef{
public static void main(String[] args)
{
    Scanner ob=new Scanner(System.in);
    int l,t;
    t=ob.nextInt();
    while(t-->0)
    {
        int a=0,b=0;
        l=ob.nextInt();
        int[] arr=new int[l];
        for(int i=0;i<l;i++)
        {
            arr[i]=ob.nextInt();
            if(i%2==1 && arr[i]%2==0) 
                    a++;
            else if(i%2==0 && arr[i]%2==1) 
                    b++;
        }
       if(a==b) 
       System.out.println(b);
       else
        System.out.println(-1);

}}}
 ```

Even Array | Editorial by manikasharma

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