Factorial trailing zeroes

Math

46.2% acceptance

Jan 12, 2026

3510

1991

Given an integer n, return the number of trailing zeroes in n!. Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1.

Solution

Rust

Time O(n)

Space O(1)

LeetCode

solution.rs

impl Solution {
  pub fn trailing_zeroes(n: i32) -> i32 {
    let mut count = 0;
    let mut power_of_five = 5;
    
    while power_of_five <= n {
      count += n / power_of_five;
      power_of_five *= 5;
    }
    
    count
  }
}