Factorial Trailing Zeroes

/*
Logic:
n!=1*2*3*4*5*6..*10*....15*....*20*...*25*....n
Trailing zeros will be only generated by 2*5,no other factors exist for 10.
so minimum of count of 2,count of 5 in factorial of n will be the answer.
but there is one cache that,2 factors we can get after every alternate number
eg 2,4,6,8,10 but 5 factor we get after every 5 numbers so obviously n! will
have more no. of 2 in their factorial so just count the 5 in n!.
after every 5 there will be one 5 do divide n by 5 but 25 have two 5 so
divide n by 25 also because numbers which are divisible by 25 will give one five
when divided by 5 and one five when divided by 25 and same goes on.

*/

class Solution {
public:
    int trailingZeroes(int n) {
        int ans=0;
        int k=5;
        while(k<=n){
            ans+=(n/k);
            k*=5;
        }
        return ans;
    }
};