Kotlin Program to Swap Two Numbers

In Kotlin, you can swap two numbers using a temporary variable like this: val temp = a; a = b; b = temp.

Examples

Inputa = 5, b = 10

OutputAfter swap: a = 10, b = 5

Inputa = -3, b = 7

OutputAfter swap: a = 7, b = -3

Inputa = 0, b = 0

OutputAfter swap: a = 0, b = 0

How to Think About It

To swap two numbers, think of holding one number temporarily so you don't lose it when you assign the other number to its place. Use a temporary container to keep one number, then replace it with the other, and finally put the saved number into the second variable.

Algorithm

1

Get the two numbers to swap.

2

Store the first number in a temporary variable.

3

Assign the second number to the first variable.

4

Assign the temporary variable's value to the second variable.

5

Print or return the swapped values.

Code

kotlin

fun main() {
    var a = 5
    var b = 10
    println("Before swap: a = $a, b = $b")
    val temp = a
    a = b
    b = temp
    println("After swap: a = $a, b = $b")
}

Output

Before swap: a = 5, b = 10
After swap: a = 10, b = 5

Dry Run

Let's trace swapping a = 5 and b = 10 through the code

1

Initial values

a = 5, b = 10

2

Store a in temp

temp = 5

3

Assign b to a

a = 10

4

Assign temp to b

b = 5

5

Final swapped values

a = 10, b = 5

Step	a	b	temp
Initial	5	10	-
Store a	5	10	5
Assign b to a	10	10	5
Assign temp to b	10	5	5

Why This Works

Step 1: Use a temporary variable

We save the value of a in temp so it is not lost when we overwrite a.

Step 2: Assign second number to first

We put the value of b into a, replacing the original a.

Step 3: Restore saved value

We assign the saved value in temp to b, completing the swap.

Alternative Approaches

Using Kotlin's built-in destructuring

kotlin

fun main() {
    var a = 5
    var b = 10
    println("Before swap: a = $a, b = $b")
    a = b.also { b = a }
    println("After swap: a = $a, b = $b")
}

This uses Kotlin's <code>also</code> function to swap without a temporary variable explicitly, but it is less readable for beginners.

Using arithmetic operations

kotlin

fun main() {
    var a = 5
    var b = 10
    println("Before swap: a = $a, b = $b")
    a = a + b
    b = a - b
    a = a - b
    println("After swap: a = $a, b = $b")
}

This swaps without extra memory but can cause overflow and is less clear.

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

Time Complexity

Swapping two numbers takes constant time because it involves a fixed number of steps without loops.

Space Complexity

Only a small temporary variable is used, so space is constant.

Which Approach is Fastest?

All approaches run in constant time; using a temporary variable is simplest and safest, while arithmetic methods save memory but risk errors.

Approach	Time	Space	Best For
Temporary variable	O(1)	O(1)	Clarity and safety
Kotlin also function	O(1)	O(1)	Concise Kotlin style
Arithmetic operations	O(1)	O(1)	Memory saving but risky