Final Challenges

You've covered the building blocks — val and var, types, conditions, loops, functions, classes, collections, null safety, and lambdas. Time to combine them.

The next two tabs are classic exercises every Kotlin developer ends up writing at some point. They both come down to a small loop and a couple of var accumulators — no new concepts, just practice.

Take your time. There's no scaffolding this time — only the function signature and an empty body.

n! is the product of every positive integer from 1 up to n:

5! = 5 × 4 × 3 × 2 × 1 = 120
3! = 3 × 2 × 1 = 6

Implement factorial(n):

• Start a var result = 1
• Multiply it by every integer from 1 to n (or n down to 1 — same answer)
• Return result

By convention, factorial(0) is 1 (the empty product).

Starter code

fun factorial(n: Int): Int {
    // Return n * (n - 1) * ... * 1
    // factorial(0) should return 1
    return 0
}

The Fibonacci sequence starts 0, 1 and each next term is the sum of the previous two:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …

The recursive definition (fib(n) = fib(n-1) + fib(n-2)) is elegant but slow. The iterative version is much faster — keep two vars, slide them forward each step:

var a = 0
var b = 1
for (i in 2..n) {
    val next = a + b
    a = b
    b = next
}
// after the loop, b is the n-th Fibonacci number (for n ≥ 1)

You'll write this in the next step.

Implement fibonacci(n) returning the n-th Fibonacci number (0-indexed):

• fibonacci(0) is 0
• fibonacci(1) is 1
• for n ≥ 2, fibonacci(n) is fibonacci(n - 1) + fibonacci(n - 2)

Use the iterative version from the previous step — handle n <= 0 and n == 1 early, then loop with two vars sliding forward.

Starter code

fun fibonacci(n: Int): Int {
    // 0, 1, 1, 2, 3, 5, 8, 13, ...
    // Return the n-th Fibonacci number (0-indexed).
    return 0
}