# How to Count Odd Numbers Below N using Java ## The challenge

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

```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

```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:

```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:

```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:

```public class OddNumbers {

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

}
```
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