Given an integer n, write a function that returns count of trailing zeroes in n!.
Input: n = 5 Output: 1 Factorial of 5 is 120 which has one trailing 0. Input: n = 20 Output: 4 Factorial of 20 is 2432902008176640000 which has 4 trailing zeroes. Input: n = 100 Output: 24
Trailing 0s in n! = Count of 5s in prime factors of n! = floor(n/5) + floor(n/25) + floor(n/125) + ....
Count of trailing 0s in 100 ! is 24
Please refer complete article on Count trailing zeroes in factorial of a number for more details!
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