The challenge

Given a number n, return the number of positive odd numbers below n, EASY!

1
2

oddCount(7) //=> 3, i.e [1, 3, 5]
oddCount(15) //=> 7, i.e [1, 3, 5, 7, 9, 11, 13]

Expect large Inputs!

Test cases

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

import org.junit.Test;
import static org.junit.Assert.assertEquals;
import org.junit.runners.JUnit4;

public class SolutionTest {
    @Test
    public void fixedTests() {
        assertEquals(7, OddNumbers.oddCount(15));
        assertEquals(7511, OddNumbers.oddCount(15023));
    }
}

The solution in Java

At first we would approach this in a programming type of way:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18

public class OddNumbers {

  public static int oddCount(int n){

    // create a variable to hold out count
    int count = 0;
    
    // loop from 1 to the value of `n`
    for(int i=1; i<n; i++) {
      // increment count if i divided by 2 has a remainder
      if (i%2!=0) count++;
    }
    
    // return out count
    return count;
  }

}

But for really large numbers, this solution will quickly timeout.

So we revisit it from a maths stand-point:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

import java.lang.*;

public class OddNumbers {

  public static int oddCount(int n) {
    // divide the number by 2 and return an integer of that
    return (int) Math.floor( n/2 );
  }

}

Or simply:

1
2
3
4
5
6
7

public class OddNumbers {

  public static int oddCount(int n) {
    return n/2;
  }

}