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172. Factorial Trailing Zeroes
Description
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
Constraints:
0 <= n <= 104
Follow up: Could you write a solution that works in logarithmic time complexity?
Solutions
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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; } }; -
class Solution: def trailingZeroes(self, n: int) -> 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; }