## Count Triplets Solution in C++

You are given an array and you need to find number of tripets of indices such that the elements at those indices are in geometric progression for a given common ratio and .

For example, . If , we have and at indices and .

**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 and , the size of and the common ratio.

The next line contains space-seperated integers .

**Constraints**

**Output Format**

Return the count of triplets that form a geometric progression.

**Sample Input 0**

```
4 2
1 2 2 4
```

**Sample Output 0**

`2`

**Explanation 0**

There are triplets in satisfying our criteria, whose indices are and

**Sample Input 1**

```
6 3
1 3 9 9 27 81
```

**Sample Output 1**

`6`

**Explanation 1**

The triplets satisfying are index , , , , and .

**Sample Input 2**

```
5 5
1 5 5 25 125
```

**Sample Output 2**

`4`

**Explanation 2**

The triplets satisfying are index , , , .

### Solution 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;
}
```