# 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;}`