Factorial Trailing Zeroes Problem


Description

LeetCode Problem 172.

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:

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

Example 2:

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

Example 3:

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Input: n = 0
Output: 0

Constraints:

  • 0 <= n <= 10^4


Sample C++ Code

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class Solution {
public:
    int trailingZeroes(int n) {
        int zeros = 0;
        uint64_t p = 5;
        while (p <= n) {
            zeros += (n / p);
            p *= 5;
        }
        return zeros;
    }
};




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