HackerEarth Divide arrays problem solution

In this HackerEarth Divide arrays problem solution You are given an array A of N integers. Find the smallest index i (1 <= i <= N – 1) such that:

mex(A[1],A[2],…,A[i]) = mex(A[i+1],A[i+2],…,A[N])

If no such index exists, print -1.

#include<bits/stdc++.h>
#define int long long int
using namespace std;

void solve(){
    int n;
    cin >> n;
    assert(1 <= n and n <= 100000);

    set<int>s;
    for(int i = 0 ; i <= n ; i++) s.insert(i);

    int a[n+1];
    for(int i = 1 ; i <= n ; i++){
        cin >> a[i];
        assert(0 <= a[i] and a[i] <= 100000);
    }

    int prefmex[n + 1];
    int sufmex[n + 1];
    for(int i = 1 ; i <= n ; i++){
        if(s.find(a[i]) != s.end())
            s.erase(a[i]);
        prefmex[i] = *(s.begin()); 
    }

    s.clear();
    for(int i = 0 ; i <= n ; i++) s.insert(i);
    for(int i = n ; i >= 1 ; i--){
        if(s.find(a[i]) != s.end())
            s.erase(a[i]);
        sufmex[i] = *(s.begin()); 
    }

    int ans = -1;
    for(int i = 1 ; i <= n - 1 ; i++){
        if(prefmex[i] == sufmex[i + 1])
        {
            cout << i << endl;
            return;
        }
    }
    cout << -1 << endl;
}

signed main(){
    int t;
    cin >> t;
    assert(1 <= t and t <= 10);
    while(t--){
        solve();
    }
}