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Problem : Given an integer n, return the number of trailing zeroes in n!.
n!
Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1.
n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1
Input: n = 5 Output: 1 Explanation: 5! = 120, one trailing zero.