Consider the following problem : Your are given an array $$$a_1, a_2, ..., a_n$$$ and q queries of form $$$l r$$$. For each query you have to find the MEX of the array $$$a_l, a_{l+1}, ..., a_r$$$. The queries are offline.

One of the approach to the problem is discussed here.

First let's consider a special case of the queries where $$$r = n$$$. In this case we want to find the smallest integer that does not occur int the subarray $$$a_l, ..., a_r$$$. Let $$$b_x = \bigl{ $$$

#include<bits/stdc++.h>
using namespace std;
#define ll long long
struct SegmentTree{
    int n;
    vector<int> tree;
    SegmentTree(int sz){
        n = 1;
        while(n < sz){
            n <<= 1;
        }
        tree.assign(2 * n, 0);
    }
    void set(int ind, int val){
        ind += n;
        tree[ind] = val;
        ind >>= 1;
        while(ind > 0){
            tree[ind] = min(tree[2 * ind], tree[2 * ind + 1]);
            ind >>= 1;
        }
    }
    int get(int x, int node = 1){
        // return the first index i, such that s[i] < x
        int ans = 1;
        while(ans < n){
            ans <<= 1;
            if(tree[ans] >= x){
                ++ans;
            }
        }
        return (ans - n);
    }
};
int main(){
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for(int i = 1; i <= n; ++i){
        cin >> a[i];
    }
    int q;
    cin >> q;
    vector<vector<pair<int, int>>> queries(n + 1);
    for(int i = 0; i < q; ++i){
        int l, r;
        cin >> l >> r;
        queries[r].push_back({l, i});
    }

    vector<int> res(q);

    SegmentTree s(n + 1);
    for(int i = 1; i <= n; ++i){
        // set the last occurence of a[i] to i
        s.set(a[i], i);

        for(auto [l, ind] : queries[i]){
            // find the smallest x, such that last occurence of x < l
            res[ind] = s.get(l);
        }
    }
    for(int elem : res){
        cout << elem << '\n';
    }
    cout << '\n';

    return 0;
}