In this **Leetcode Factorial Trailing Zeroes problem solution** we have Given an integer n, return the number of trailing zeroes in n!.

## Problem solution in Python.

def trailingZeroes(self, n): index = 1 total_power_of_five = 0 while n // (5**index) > 0: total_power_of_five += n // (5**index) index += 1 return total_power_of_five

## Problem solution in Java.

class Solution { public int trailingZeroes(int n) { int div5 = 0; long pow5 = 5; while (pow5 <= n) { div5 += n/pow5; pow5 *= 5; } return div5; } }

## Problem solution in C++.

class Solution { public: int trailingZeroes(int n) { int base = 5; int count = 0; while (n / base) { count += (n / base); n /= base; } return count; } };

## Problem solution in C.

int trailingZeroes(int n) { int count = 0; int base = 5; for (; n != 0; n /= base) count += n / base; return count; }

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