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16.05. Factorial Zeros

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

Thinking

Trailing zeros of \(n!\) come from factors \(10=2\times 5\). Twos are plentiful, so the count equals the number of fives in \([1,n]\), including higher powers.

Computing \(n!\) and counting zeros overflows for large \(n\).

Repeated \(n//=5\) adds the contributions of \(5\), \(5^2\), \(5^3,\ldots\) in \(O(\log n)\) time.

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 \(O(1)\).

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

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

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

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

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

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class Solution {
    func trailingZeroes(_ n: Int) -> Int {
        var count = 0
        var number = n
        while number > 0 {
            number /= 5
            count += number
        }
        return count
    }
}

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