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--;
        }
	}
}

There are 2 approaches to solve this problem.

DataStructure approach

We use a dequeue, it stands for a double ended queue. Use it to keep on removing the first or last element and maintain to variables for each of the player. repeat this process till the deque is empty.

2 pointer approach

Same as the above approach instead of the data structure, we just keep to pointers at the start and the end and do the same process.

// 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 n;
    cin >> n;
    deque<ll> a(n);
    for (int i = 0; i < n; i++) { cin >> a[i]; }
    ll one = 0, two = 0;
    while (!a.empty())
    {
        if (a.front() > a.back())
        {
            one += a.front();
            a.pop_front();
        }
        else
        {
            one += a.back();
            a.pop_back();
        }
        if (!a.empty())
        {
            if (a.front() > a.back())
            {
                two += a.front();
                a.pop_front();
            }
            else
            {
                two += a.back();
                a.pop_back();
            }
        }

    }
    cout << one << " " << two;
    nl;

    return 0;
}

```

```

4 hours = 240 minutes
time taken to get the party = x
so, 240-x is the time we have for solving problems.

// 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 n, k;
    cin >> n >> k;
    int x = (240 - k);
    ll ans = -1;
    for (int i = 1; i <= n; i++)
    {
        if (x >= 5 * i)
        {
            x -= 5 * i;
            ans = i;
        }
        else { break; }
    }
    cout << ans;
    nl;
    return 0;
}

```

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