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. The maintain 2 pointers the left pointer will be looking for the first 1 and the right pointer will be looking for the first 0. if they both don't cross each other, then simply replace. loop ends if left pointer becomes greater than right pointer or they become equal.

// Aur Bhai Dekhne aagaye ;)
// Author: Abhijit Mishra 
#include <bits/stdc++.h>
using namespace std;


#define nl cout << "\n"
#define all(x) x.begin(), x.end()
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);


typedef long long int ll;
typedef vector<ll> vi;


int main()
{
    fast;
    ll t = 1;
    cin >> t;
    while (t--)
    {
        ll n;
        cin >> n;
        vi a(n);
        for (int i = 0; i < n; i++)
        {
            cin >> a[i];
        }

        if (is_sorted(all(a))) { cout << 0; }
        else
        {
            ll ans = 0;
            ll l = 0, r = n - 1;
            while (l != r)
            {
                if (a[l] == 0) { l++; }
                else if (a[r] == 1) { r--; }
                else if (a[l] == 1 && a[r] == 0)
                {
                    a[l] = 0;
                    a[r] = 1;
                    ans++;
                }
            }
            if (is_sorted(all(a))) { cout << ans; }
        }
        nl;

    }
    return 0;
}

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--;
        }
	}
}