Given an integer n, write a function that returns count of trailing zeroes in n!. Examples :
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) + ....
C++
// C++ program to count // trailing 0s in n! #include <iostream> using namespace std;
// Function to return trailing // 0s in factorial of n int findTrailingZeros( int n)
{ // Initialize result
int count = 0;
// Keep dividing n by powers of
// 5 and update count
for ( int i = 5; n / i >= 1; i *= 5)
count += n / i;
return count;
} // Driver Code int main()
{ int n = 100;
cout << "Count of trailing 0s in " << 100
<< "! is " << findTrailingZeros(n);
return 0;
} |
Output:
Count of trailing 0s in 100! is 24
Time Complexity: O(log5n)
Auxiliary Space: O(1)
Please refer complete article on Count trailing zeroes in factorial of a number for more details!
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