# Factorial Trailing Zeroes LeetCode Solution

Here, We see Factorial Trailing Zeroes LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

Given an integer `n`, return the number of trailing zeroes in `n!`.

Note that `n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1`.

Example 1:
Input: n = 3
Output: 0
Explanation: 3! = 6, no trailing zero.

Example 2:
Input: n = 5
Output: 1
Explanation: 5! = 120, one trailing zero.

Example 3:
Input: n = 0
Output: 0

## Factorial Trailing Zeroes LeetCode SolutionC++

``````class Solution {
public:
int trailingZeroes(int n) {
int count = 0;
for (long long i = 5; n / i; i *= 5)
count += n / i;
return count;
}
};```Code language: PHP (php)```

## Factorial Trailing Zeroes SolutionJava

``````class Solution {
public int trailingZeroes(int n) {
int r = 0;
while (n > 0) {
n /= 5;
r += n;
}
return r;
}
}```Code language: PHP (php)```

## Factorial Trailing Zeroes SolutionJavaScript

``````var trailingZeroes = function(n) {
let numZeroes = 0;
for (let i = 5; i <= n; i *= 5) {
numZeroes += Math.floor(n / i);
}
return numZeroes;
};```Code language: JavaScript (javascript)```

## Factorial Trailing Zeroes SolutionPython

``````class Solution(object):
def trailingZeroes(self, n):
if(n < 0):
return -1
output = 0
while(n >= 5):
n //= 5
output += n
return output ```Code language: HTML, XML (xml)```
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