for query you can use sparse table or segment tree

I solved it this way: https://github.com/actium/cf/blob/master/edu/09/2g.cpp

In tutorial, we can easily get max/min when adding an element, but it's hard to remove an element.

How does the tutorial solve it?

It uses two stacks to implement the queue, and get the result of max/min from two of the stack to get the result.

Can we just replace it with other merge function?

#include <bits/stdc++.h>
#define pb push_back
#define ppb pop_back
#define ll long long 
using namespace std;
 
class Stack {
    public:
    vector<ll> vals, gcds;
 
    void push(ll a){
        ll cur;
        if(gcds.empty()) cur=a;
        else cur = __gcd(gcds.back(), a);
        vals.pb(a);
        gcds.pb(cur);
    }
 
    ll pop(){
        ll x = vals.back();
        vals.ppb();
        gcds.ppb();
        return x;
    }
 
    ll gcd(){
        return gcds.back();
    }
 
    ll val() {
        return vals.back();
    }
} s1,s2;
 
void remove(){
    if(s2.vals.empty()){
        while(!s1.vals.empty()) s2.push(s1.pop());
    }
    s2.pop();
}
 
ll tot_gcd(){
    if(s1.gcds.empty() && s2.gcds.empty()) return -1;
    if(s1.gcds.empty()) return s2.gcd();
    if(s2.gcds.empty()) return s1.gcd();
    return gcd(s1.gcd(), s2.gcd());
}
 
void solve()
{
    ll n;
    cin>>n;
 
    vector<ll>v(n);
    for(auto &it : v) cin>>it;
 
    ll l = 0;
    ll ans = INT_MAX;
    for(auto r=0;r<n;r++){
        s1.push(v[r]);
        while(tot_gcd() == 1){
            ans = min(ans, r-l+1);
            remove();
            l++;
        }
    }
 
    if(ans == INT_MAX) cout<<-1<<nline;
    else cout<<ans<<nline;
}

This was my solution, I suggest that you read through the theory in step 2 segment with small spread. It's sad to see that there is no proper editorial for the problems in edu.