Kotlin Program to Check Even or Odd Number

In Kotlin, you can check if a number is even or odd using if (number % 2 == 0) to test evenness and print the result accordingly.

Examples

Input4

Output4 is even

Input7

Output7 is odd

Input0

Output0 is even

How to Think About It

To check if a number is even or odd, you divide it by 2 and look at the remainder. If the remainder is 0, the number is even; otherwise, it is odd. This is like splitting candies into pairs and seeing if any candy is left alone.

Algorithm

1

Get the input number

2

Calculate the remainder when the number is divided by 2

3

If the remainder is 0, the number is even

4

Otherwise, the number is odd

5

Print the result

Code

kotlin

fun main() {
    val number = 7
    if (number % 2 == 0) {
        println("$number is even")
    } else {
        println("$number is odd")
    }
}

Output

7 is odd

Dry Run

Let's trace the number 7 through the code to see how it checks even or odd.

1

Input number

number = 7

2

Calculate remainder

7 % 2 = 1

3

Check remainder

Since remainder is 1 (not 0), number is odd

4

Print result

Output: "7 is odd"

Step	Operation	Value
1	Input number	7
2	7 % 2	1
3	Check if remainder == 0	False
4	Print output	"7 is odd"

Why This Works

Step 1: Using modulus operator

The % operator gives the remainder of division, which helps us find if a number divides evenly by 2.

Step 2: Even number check

If the remainder is 0, it means the number is divisible by 2 and is even.

Step 3: Odd number check

If the remainder is not 0, the number is not divisible by 2 and is odd.

Alternative Approaches

Using when expression

kotlin

fun main() {
    val number = 10
    when (number % 2) {
        0 -> println("$number is even")
        else -> println("$number is odd")
    }
}

This uses Kotlin's <code>when</code> expression for cleaner branching but works the same way.

Using a function to return boolean

kotlin

fun isEven(num: Int): Boolean = num % 2 == 0

fun main() {
    val number = 3
    if (isEven(number)) println("$number is even") else println("$number is odd")
}

This separates logic into a function, making code reusable and clearer.

Complexity: O(1) time, O(1) space

Time Complexity

The operation uses a single modulus calculation and a simple condition, so it runs in constant time.

Space Complexity

No extra memory is needed besides the input number and a few variables, so space is constant.

Which Approach is Fastest?

All approaches use the modulus operator once, so they have the same speed; differences are mainly in code style.

Approach	Time	Space	Best For
If-else with modulus	O(1)	O(1)	Simple and clear checks
When expression	O(1)	O(1)	Cleaner branching syntax
Function returning boolean	O(1)	O(1)	Reusable logic in larger programs