# [front end algorithm] zero after factorization, twice traversal

Time：2021-6-6

### Given an integer n, return n! The number of zeros in the result mantissa.

``````Example 1:
Input: 3
Output: 0
Explanation: 3= 6, there is no zero in the mantissa.``````
``````Example 2:
Input: 5
Output: 1
Explanation: 5= 120, one zero in the mantissa``````
• Note: the time complexity of your algorithm should be o (log n).

Solution code:

``````/**
* @param {number} n
* @return {number}
*/
var trailingZeroes = function(n) {
let result = 1;
let i = 1;
while (i <= 5) {
result *= i;
i++;
};
var num = 0;
while (n > 0) {
n = (n - (n % 5)) / 5;
num += n;
}
return num
};``````

Knowledge points

• Find the zero after the factorial number

The number of zeros at the end depends on the number of factors 2 and 5 in the multiplication. Obviously, the number of factors 2 in multiplication is greater than that of 5, so we only need to count the number of factors 5.

``````Such as 5!
Please 5! 5/5 = 1；// 120 followed by a zero``````
``````The number of multiples of 5 is 1024 / 5 = 204
The number of multiples of 25 is 1024 / 25 = 40
The number of multiples of 125 is: 1024 / 125 = 8
The number of multiples of 625 is 1024 / 625 = 1
So 1024! There are 204 + 40 + 8 + 1 = 253 factor 5.
That is to say, 1024! There are 253 zeros at the end.``````

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