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Description:
Given an integer $n$ , return the number of trailing zeros in $n!$ (n factorial).
Factorial is defined as:
$$
n! = n \times (n – 1) \times (n – 2) \times \dots \times 3 \times 2 \times 1
$$
Examples:
Example 1:
Input: n = 3
Output: 0
Explanation: 3! = 6, which has no trailing zeros.
Example 2:
Input: n = 5
Output: 1
Explanation: 5! = 120, which has one trailing zero.
Example 3:
Input: n = 0
Output: 0
Constraints:
- $0 \leq n \leq 10^4$
class Solution {
public:
int trailingZeroes(int n) {
int count;
while(n >= 5){
n = n / 5;
count += n;
}
return count;
}
};
/*
## Counting 5 as a factor.
We Need To Determine how many times "10" appears as a factor in the product.
10 is formed by multiplying 2 & 5, we can only count 5 as a factor.
For example:
- 3! = 3*2*1 = 6 -> +0
- 5! = 5*4*3*2*1 = 120 -> +1
- 10! = 10*9*...*3*2*1 = 3628800 -> +2
*/
EN 
CN
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