Geek For Geeks | Trailing Zeroes in a factorial of a number | Java
For an integer N find the number of trailing zeroes in N!. Example 1: Input: N = 5 Output: 1 Explanation: 5! = 120 so the number of trailing zero is 1. Expected Time Complexity: O(logN) Expected Auxiliary Space: O(1) Constraints: 1 <= N <= 10 9 public class TrailingZeroes { public static void main (String[] args) { int n = 384 ; int count = 0 ; for ( int i = 5 ; n/i >= 1 ; i *= 5 ) { count += n/i ; } System. out .println(count) ; } }