Kotlin Program to Find Largest Number in a List

You can find the largest number in a list in Kotlin using list.maxOrNull(), for example: val largest = list.maxOrNull().

Examples

Input[3, 5, 1, 9, 2]

Output9

Input[10, 10, 10]

Output10

Input[]

Outputnull

How to Think About It

To find the largest number in a list, look at each number one by one and remember the biggest number you have seen so far. At the end, the number you remembered is the largest.

Algorithm

1

Get the list of numbers.

2

Start with the first number as the largest.

3

Compare each number in the list with the current largest number.

4

If a number is bigger, update the largest number.

5

After checking all numbers, return the largest number.

Code

kotlin

fun main() {
    val numbers = listOf(3, 5, 1, 9, 2)
    val largest = numbers.maxOrNull()
    println(largest)
}

Output

9

Dry Run

Let's trace the list [3, 5, 1, 9, 2] through the code.

1

Initialize list

numbers = [3, 5, 1, 9, 2]

2

Find max

largest = numbers.maxOrNull() -> 9

3

Print result

Print 9

Iteration	Current Number	Current Largest
1	3	3
2	5	5
3	1	5
4	9	9
5	2	9

Why This Works

Step 1: Use maxOrNull() function

The maxOrNull() function checks all elements and returns the largest one or null if the list is empty.

Step 2: Handle empty list

If the list is empty, maxOrNull() returns null, so the program won't crash.

Step 3: Print the largest number

The program prints the largest number found in the list.

Alternative Approaches

Manual loop to find largest

kotlin

fun main() {
    val numbers = listOf(3, 5, 1, 9, 2)
    var largest = numbers[0]
    for (num in numbers) {
        if (num > largest) {
            largest = num
        }
    }
    println(largest)
}

This method uses a loop to find the largest number manually, which is more verbose but shows the logic clearly.

Using reduce function

kotlin

fun main() {
    val numbers = listOf(3, 5, 1, 9, 2)
    val largest = numbers.reduce { max, current -> if (current > max) current else max }
    println(largest)
}

This uses <code>reduce</code> to compare elements pairwise, which is concise but requires the list to be non-empty.

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

Time Complexity

The program checks each element once, so it takes time proportional to the number of elements, which is O(n).

Space Complexity

It uses only a few variables regardless of list size, so space is O(1).

Which Approach is Fastest?

Using maxOrNull() is efficient and concise; manual loops and reduce have similar performance but more code.

Approach	Time	Space	Best For
maxOrNull()	O(n)	O(1)	Simple and safe for any list
Manual loop	O(n)	O(1)	Learning and understanding logic
reduce()	O(n)	O(1)	Concise code but non-empty lists only