Little Elephant and Subarrays O(N) Solution ??

i tried to implement the O(n) solution but couldn’t , here is the editorial https://discuss.codechef.com/problems/SUBMIN

I tried to implement the O(n) solution but wasn’t able to , here is the code and link .

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <stack>
#include <list>
typedef long long int ll;
using namespace std;
ll a[1000000+100],b[1000000+100]={0};
int main() {
	// your code goes here
	ll n;
	cin>>n;
	for(ll i=0;i<n;i++)cin>>a[i];
	list<ll> k1,k2;
	for(ll i=0,j=n-1;i<n;i++,j--)
	{
		ll k11,k22;
		while(!k1.empty()&&a[k1.back()]>a[i])
		k1.pop_back();
		while(!k2.empty()&&a[k2.back()]>a[j])
		k2.pop_back();
		if(k1.empty())k11=0;
		else k11 = 1+k1.back();
		if(k2.empty())k22=n-1;
		else k22 = k2.back()+1;
		b[a[i]] += (i-k11+1)*(k22-i+1);
		k1.push_back(i);
		k2.push_back(i);
		
	}
	ll q;
	cin>>q;
	while(q--)
	{
		ll temp;
		cin>>temp;
		cout<<b[temp]<<endl;
	}
	return 0;
}
//