HackerEarth A Game of Numbers problem solution

In this HackerEarth A Game of Numbers problem solution, You are given an array A of N integers. Now, two functions F(X) and G(X) are defined:

1. F(X) : This is the smallest number Z such that X < Z <= N and A[X] < A[Z]
2. G(X) : This is the smallest number Z such that X < Z <= N and A[X] > A[Z]

Now, you need to find for each index i of this array G(F(i)), where 1 <= i <= N. If such a number does not exist, for particular index i, output 1 as its answer. If such a number does exist, output A[G(F(i))]

HackerEarth A Game of Numbers problem solution.

#include<bits/stdc++.h>
using namespace std;
 
int main()
{

    int n;
    cin>>n;
    
    short int a[n];
    for(int i=0;i<n;i++)cin>>a[i];
    short int next_greater[n], next_smaller[n];
    stack<short int> s1;
    for (short int i=n-1; i>=0; i--)
    {
        while (!s1.empty() && a[s1.top()] <= a[i] )
            s1.pop();
        if (!s1.empty())
            next_greater[i] = s1.top();
        else
            next_greater[i] = -1;

        s1.push(i);
    }

    stack<short int> s2;
    for (short int i=n-1; i>=0; i--)
    {
        while (!s2.empty() && a[s2.top()] >= a[i])
            s2.pop();
        if (!s2.empty())
            next_smaller[i] = s2.top();

        else
            next_smaller[i] = -1;

        s2.push(i);
    }
    
    
    for (short int i=0; i< n; i++)
    {

        if (next_greater[i] != -1 && next_smaller[next_greater[i]] != -1)
            cout << a[next_smaller[next_greater[i]]] <<" ";
        else
            cout<<-1<<" ";
    }
   
    return 0;
}