Number of trailing zeros of N in Java

The challenge

Write a program that will calculate the number of trailing zeros in a factorial of a given number.

N! = 1 * 2 * 3 * ... * N

Be careful 1000! has 2568 digits…

For more info, see: http://mathworld.wolfram.com/Factorial.html

Examples

zeros(6) = 1
// 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720 --> 1 trailing zero

zeros(12) = 2
// 12! = 479001600 --> 2 trailing zeros

The solution in Java code

Option 1:

public class Solution {
  public static int zeros(int n) {
    int res = 0;
    for (int i = 5; i <= n; i *= 5) {
      res += n / i;
    }
    return res;
  }
}

Option 2:

public class Solution {
  public static int zeros(int n) {
      if(n/5 == 0)
        return 0;
      return n/5 + zeros(n/5);
  }
}

Option 3:

public class Solution {
  public static int zeros(final int n) {
    return (n < 5) ? 0 : (n / 5) + zeros(n / 5);
  }
}

Test cases to validate our solution

import org.junit.Test;
import static org.hamcrest.CoreMatchers.*;
import static org.junit.Assert.assertThat;

public class SolutionTest {
  @Test
  public void testZeros() throws Exception {
    assertThat(Solution.zeros(0), is(0)); 
    assertThat(Solution.zeros(6), is(1)); 
    assertThat(Solution.zeros(14), is(2));    
  }
}
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