"""
Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.
"""

@@ -124,12 +124,227 @@
 lexity.%0A%22%22%22%0A
+%0A%0Aclass Solution:%0A    # @param %7Binteger%7D n%0A    # @return %7Binteger%7D%0A    def trailingZeroes(self, n):%0A        x = 5%0A        ans = 0%0A        while n %3E= x:%0A            ans += n / x%0A            x *= 5%0A        return ans%0A

Add Factorial Trailing Zeroes.