Since only entries in array are binary digits, all 0s must be before 1s. So we can use two pointers to interchange 0s to 1s and count the operations.

[submission:185388441]

import java.util.*;


public class Rebellion {
	public static void main(String[]args){

        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t>0){
            int n = sc.nextInt();
            int[] arr = new int[n];
            for(int i = 0;i<n;i++) arr[i] = sc.nextInt();
            int p1 = 0;
            int p2 = n-1;
            int ans = 0;
            while(p1<p2){
                if(arr[p1] == 0){
                    p1++;
                    continue;
                }
                if(arr[p2] == 1){
                    p2--;
                    continue;
                }
                if(arr[p1] ==  1 && arr[p2] ==  0){
                    ans++;
                    p1++;
                    p2--;
                }
            }
            System.out.println(ans);
            t--;
        }
	}
}

[submission:185389438]

[submission:185390659]

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[submission:185393751]

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[submission:185394097]

[submission:185426903]

[submission:185394223]