hackerrank-count-triplets-1.py

from collections import Counter

# Imagine this as three boxes or stages, each number "graduates" from the first stage to the second one, then, to the final stage.
# Instead of having a final stage object, having an integer that's incremented each time a number reaches the final stage, is more performant

def countTriplets(arr, r):
    finalStage = 0
    firstStage = Counter()
    secondStage = Counter()
    for i in reversed(arr):
        finalStage += secondStage[i*r]
        secondStage[i] += firstStage[i*r]
        firstStage[i] += 1
    return finalStage

n, r = [int(x) for x in input().strip().split()]
numbers = [int(x) for x in input().strip().split()]

print(countTriplets(numbers, r))