This is the question for which the implementation is mentioned below

Basics of Implementation & Basic Programming Practice Problems

#include<bits/stdc++.h>
using namespace std;
 
const int INF = 1e9;
const int N = 100031;
 
int n, tests;
int level[N];
vector<pair<int, int> > v;
int ans[N];
 
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> tests;
 
    for (int i = 0; i < n; i++)
    {
	cin >> level[i];
    }
    v.clear();
    for (int i = 0; i < n; i++)
    {
	v.push_back(make_pair(level[i], i));
    }
    for (int i = 0; i < n; i++)
	ans[i] = 0;
    while (v.size()>1)
    {
	vector<pair<int, int> > v2;
        for (int i = 0; i < v.size(); i += 2)
        {
            if (i + 1 == v.size())
            {
		v2.push_back(v[i]);
                continue;
            }
	    ans[v[i].second]++;
	    ans[v[i + 1].second]++;
            if (v[i].first>v[i + 1].first)
		v2.push_back(v[i]);
            else
		v2.push_back(v[i + 1]);
        }
	v = v2;
    }
    for (; tests; --tests)
    {
        int val;
        cin >> val;
        --val;
	cout << ans[val] << "\n";
    }
    return 0;
}

---------

I'm not sure about the complexity of this code. I believe it's O(log(n)^2).

Or is it just O(n), as N elements are input in the first loop at the beginning

Let me explain my rationale. The loop

(while (v.size()>1) { .. } ) 

runs log(n)[ explanation.. first there are N then N/2 , then N/4 elements .. till there is just 1 element left ] so that's a complexity of log(N) ,

However the inner loop too runs in log(n) time, each time, but n is variable each time.

vector<pair<int, int> > v2;
for (int i = 0; i < v.size(); i += 2)
{
    if (i + 1 == v.size())
    {
         v2.push_back(v[i]);
         continue;
    }
    ans[v[i].second]++;
    ans[v[i + 1].second]++;
    if (v[i].first>v[i + 1].first)
        v2.push_back(v[i]);
    else
        v2.push_back(v[i + 1]);
}
v = v2;