HackerRank Count Triplets problem solution

In this HackerRank Count Triplets Interview preparation kit problem solution You are given an array and you need to find a number of triplets 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.

Problem solution in Python programming.

#!/bin/python3

import math
import os
import random
import re
import sys
from collections import Counter

# Complete the countTriplets function below.
def countTriplets(arr, r):
    a = Counter(arr)
    b = Counter()
    s = 0
    for i in arr:
        j = i//r
        k = i*r
        a[i]-=1
        if b[j] and a[k] and not i%r:
            s+=b[j]*a[k]
        b[i]+=1
    return s

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    nr = input().rstrip().split()

    n = int(nr[0])

    r = int(nr[1])

    arr = list(map(int, input().rstrip().split()))

    ans = countTriplets(arr, r)

    fptr.write(str(ans) + 'n')

    fptr.close()

Problem solution in Java Programming.

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

public class Solution {

    // Complete the countTriplets function below.
    private static long countTriplets(List<Long> arr, long r) {
        Map<Long, Long> potential = new HashMap<>();
        Map<Long, Long> counter = new HashMap<>();
        long count = 0;
        for (int i = 0; i < arr.size(); i++) {
            long a = arr.get(i);
            long key = a / r;
            
            if (counter.containsKey(key) && a % r == 0) {
                count += counter.get(key);
            }
            
            if (potential.containsKey(key) && a % r == 0) {
                long c = potential.get(key);
                counter.put(a, counter.getOrDefault(a, 0L) + c);
            }
            
            potential.put(a, potential.getOrDefault(a, 0L) + 1); // Every number can be the start of a triplet.
        }
        return count;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        String[] nr = bufferedReader.readLine().replaceAll("\s+$", "").split(" ");

        int n = Integer.parseInt(nr[0]);

        long r = Long.parseLong(nr[1]);

        List<Long> arr = Stream.of(bufferedReader.readLine().replaceAll("\s+$", "").split(" "))
            .map(Long::parseLong)
            .collect(toList());

        long ans = countTriplets(arr, r);

        bufferedWriter.write(String.valueOf(ans));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Problem solution in C++ programming.

#include <bits/stdc++.h>

using namespace std;

// Complete the countTriplets function below.
typedef long long ll;

int main()
{
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);

    long n,r;
    cin >> n >> r;

    map<int,long> mp2, mp3;
    //mp2 to hold count of needed values after this one to complete 
    //2nd part of triplet
    //mp3 to hold count of needed values to complete triplet

    int val;
    long long res = 0;
    while(n--)
    {
        cin >> val;
        if (mp3.count(val))        //This value completes mp3[val] triplets
            res += mp3[val];
        if (mp2.count(val))        //This value is valid as 2° part of mp2[val] triplets
            mp3[val*r] += mp2[val];
        mp2[val*r]++;            //"Push-up" this value as possible triplet start
    }

    cout << res << endl;

    return 0;
}