# Simple C/C++ Solution (with detailed explaination)

• The idea is:

1. The ZERO comes from 10.
2. The 10 comes from 2 x 5
3. And we need to account for all the products of 5 and 2. likes 4×5 = 20 ...
4. So, if we take all the numbers with 5 as a factor, we'll have way more than enough even numbers to pair with them to get factors of 10

Example One

How many multiples of 5 are between 1 and 23? There is 5, 10, 15, and 20, for four multiples of 5. Paired with 2's from the even factors, this makes for four factors of 10, so: 23! has 4 zeros.

Example Two

How many multiples of 5 are there in the numbers from 1 to 100?

because 100 ÷ 5 = 20, so, there are twenty multiples of 5 between 1 and 100.

but wait, actually 25 is 5×5, so each multiple of 25 has an extra factor of 5, e.g. 25 × 4 = 100，which introduces extra of zero.

So, we need know how many multiples of 25 are between 1 and 100? Since 100 ÷ 25 = 4, there are four multiples of 25 between 1 and 100.

Finally, we get 20 + 4 = 24 trailing zeroes in 100!

The above example tell us, we need care about 5, 5×5, 5×5×5, 5×5×5×5 ....

Example Three

By given number 4617.

5^1 : 4617 ÷ 5 = 923.4, so we get 923 factors of 5

5^2 : 4617 ÷ 25 = 184.68, so we get 184 additional factors of 5

5^3 : 4617 ÷ 125 = 36.936, so we get 36 additional factors of 5

5^4 : 4617 ÷ 625 = 7.3872, so we get 7 additional factors of 5

5^5 : 4617 ÷ 3125 = 1.47744, so we get 1 more factor of 5

5^6 : 4617 ÷ 15625 = 0.295488, which is less than 1, so stop here.

Then 4617! has 923 + 184 + 36 + 7 + 1 = 1151 trailing zeroes.

C/C++ code

``````int trailingZeroes(int n) {
int result = 0;
for(long long i=5; n/i>0; i*=5){
result += (n/i);
}
return result;
}
``````

---------update-----------

To avoid the integer overflow as @localvar mentioned below(in case of 'n >=1808548329' ), the expression " i <= INT_MAX/5" is not a good way to prevent overflow, because 5^13 is > INT_MAX/5 and it's valid.

So, if you want to use "multiply", consider define the 'i' as 'long long' type.

Or, take the solution @codingryan mentioned in below answer!

• This post is deleted!

• This post is deleted!

• ``````int trailingZeroes(int n) {
int sum=0;
int tmp=0;
while(n/5>0)
{
tmp=n/5;
sum+=tmp;
n=tmp;
}
return sum;
}
``````

explanation:http://www.purplemath.com/modules/factzero.htm

• This post is deleted!

• +1. Excellent analysis and very detailed thought process. Welcome to Discuss, @haoel!

• good explaination, Hao. but there's a small mistake, will get wrong answer when 'n >=1808548329' because of integer overflow. And leetcode are currently lack this case

• Yes, you are right. in your case, we need check the integer overflow in for-loop: i <= INT_MAX/5

• still problem because 5^13 > INT_MAX/5 and valid, maybe the best way is to use divide instead of multiply

• Oh, yes! if we need to use multiply, it seems we have to use "long long".

• Kudos for detailed analysis and expansion with generous examples.

• @localvar: Excellent observation. I have just added your suggested test case. Thanks!

• optimized to 3ms:

int trailingZeroes(int n)
{
int sum=0;

``````while(n>0)
{
n=n/5;
sum+=n;
}

return sum;
``````

}

• excellent! How do you come up with this solution?

• Great explanation!! It's so easy to understand!! I just add the python version.

``````def trailingZeroes(self, n):
# return n/5 + n/25 + n/125 + ... + n/(5^13)
return sum([ n/(5**i) for i in range(1, 14) ])``````

• actually your solution is 8 ms @ engkfke

this is optimized

``for(int res=0;;res+=(n/=5)) if(n==0) return res;``

• I remember I've seen your article years ago. Are you work at Alibaba right now? // off the topic.

• Thank you for your explanation!

• Really nice explanation, I forgot the situation of extra '5' induced by numbers like 25, appreciate it.

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