Question

16.05. Factorial Zeros

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Description

Write an algorithm which computes the number of trailing zeros in n factorial.

Example 1:

Input: 3

Output: 0

Explanation: 3! = 6, no trailing zero.

Example 2:

Input: 5

Output: 1

Explanation: 5! = 120, one trailing zero.

Note: Your solution should be in logarithmic time complexity.

Solutions

Solution 1: Mathematics

The problem is actually asking for the number of factors of $5$ in $[1,n]$.

Let's take $130$ as an example:

1. Divide $130$ by $5$ for the first time, and get $26$, which means there are $26$ numbers containing a factor of $5$.
2. Divide $26$ by $5$ for the second time, and get $5$, which means there are $5$ numbers containing a factor of $5^2$.
3. Divide $5$ by $5$ for the third time, and get $1$, which means there is $1$ number containing a factor of $5^3$.
4. Add up all the counts to get the total number of factors of $5$ in $[1,n]$.

The time complexity is $O(\log n)$, and the space complexity is $O(1)$.

class Solution:
  def trailingZeroes(self, n: int) -> int:
    ans = 0
    while n:
      n //= 5
      ans += n
    return ans

class Solution {
  public int trailingZeroes(int n) {
    int ans = 0;
    while (n > 0) {
      n /= 5;
      ans += n;
    }
    return ans;
  }
}

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

func trailingZeroes(n int) int {
  ans := 0
  for n > 0 {
    n /= 5
    ans += n
  }
  return ans
}

function trailingZeroes(n: number): number {
  let ans = 0;
  while (n > 0) {
    n = Math.floor(n / 5);
    ans += n;
  }
  return ans;
}