**Count Triplets**

### Problem Statement :

You are given an array and you need to find number of tripets of indices (i, j, k) such that the elements at those indices are in geometric progression for a given common ratio r and i < j < k . Function Description Complete the countTriplets function in the editor below. It should return the number of triplets forming a geometric progression for a given as an integer. countTriplets has the following parameter(s): arr: an array of integers r: an integer, the common ratio Input Format The first line contains two space-separated integers n and r, the size of arr and the common ratio. The next line contains n space-seperated integers arr[i] . Constraints 1 <= n <= 10^5 1 <= r <= 10^9 1 <= arr[i] <= 10^9 Output Format Return the count of triplets that form a geometric progression.

### Solution :

` ````
Solution in C++ :
In C++ :
#include<bits/stdc++.h>
using namespace std;
map<long long,long long> l,r;
int main()
{
long long n, k,ans=0 ;
scanf("%lld%lld",&n,&k);
long long a[n];
for(int i=0;i<n;i++)
scanf("%lld",&a[i]);
for(int i=0;i<n;i++)
r[a[i]]++;
for(int i=0;i<n;i++)
{
r[a[i]]--;
if(a[i]%k==0)
{
ans+=l[a[i]/k]*r[a[i]*k];
}
l[a[i]]++;
}
printf("%lld\n",ans);
return 0;
}
```

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