Since \(0,1,\ldots,N\) has \(N+1\) integers, the upper bound of the answer is \(N\).

For each \(ans=0,1,2,\ldots,N\), solve the following problem:

Does \(A\) contains \(ans\)?

Once an integer that does not contained in \(A\) for the first time, we can determine that that is the answer.

The time complexity is \(O(N^2)\).

With a boolean array, the problem can also be solved in a total of \(O(N)\) too.

Sample code in \(O(N^2)\)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

signed main(){
    ll n;cin>>n;
    vector<ll>a(n);
    for(ll i=0;i<=n-1;i++)cin>>a[i];
    for(ll ans=0;ans<=n;ans++){
        bool ok=true;
        for(ll x:a){
            if(x==ans)ok=false;
        }
        if(ok){
            cout<<ans<<endl;
            return 0;
        }
    }
    return 0;
}