Count Triplets hackerRank Solution: Looking for Count Triplets solution for Hackerrank problem? Get solution with source code and detailed explainer video.
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).
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Explanation Video:

Youtube Channel partner: CodingCart
Mentor: Satyendra Jaiswal
Langauge: Python

Source Code:  for countTriplets function


# Complete the countTriplets function below.
def countTriplets(arr, r):
    count=0
    bef={}
    aft={}
    for v in arr:
        if v in aft:
            aft[v]+=1
        else:
            aft[v]=1   
    for v in arr:
        aft[v]-=1  #current      
        if v//r in bef and v %r ==0 and v*r in aft:
            count+=bef[v//r]*aft[v*r]
        if v in bef:
            bef[v]+=1
        else:
            bef[v]=1  
    return count
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Full Source Code: countTriplets HackerRank Solution


#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the countTriplets function below.
def countTriplets(arr, r):
    count=0
    bef={}
    aft={}
    for v in arr:
        if v in aft:
            aft[v]+=1
        else:
            aft[v]=1   
    for v in arr:
        aft[v]-=1  #current      
        if v//r in bef and v %r ==0 and v*r in aft:
            count+=bef[v//r]*aft[v*r]
        if v in bef:
            bef[v]+=1
        else:
            bef[v]=1  
    return count              


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()
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