Kotlin Program to Remove Duplicates from a List

You can remove duplicates in Kotlin by converting a list to a set using toSet() or by using distinct() on the list, for example: val uniqueList = list.distinct().

Examples

Input[1, 2, 2, 3, 4, 4, 5]

Output[1, 2, 3, 4, 5]

Input["apple", "banana", "apple", "orange"]

Output["apple", "banana", "orange"]

Input[]

Output[]

How to Think About It

To remove duplicates, think about keeping only one copy of each item. Since duplicates are repeated items, you can use a collection that automatically ignores repeats, like a set, or use a function that filters out repeated elements while keeping the original order.

Algorithm

1

Get the input list with possible duplicates.

2

Use a method to filter out repeated items, such as converting to a set or using a distinct function.

3

Return or print the new list with duplicates removed.

Code

kotlin

fun main() {
    val list = listOf(1, 2, 2, 3, 4, 4, 5)
    val uniqueList = list.distinct()
    println(uniqueList)
}

Output

[1, 2, 3, 4, 5]

Dry Run

Let's trace the list [1, 2, 2, 3, 4, 4, 5] through the code using distinct()

1

Original list

[1, 2, 2, 3, 4, 4, 5]

2

Apply distinct()

Scans list and keeps first occurrence of each element

3

Resulting list

[1, 2, 3, 4, 5]

Index	Element	Is Duplicate?	Action
0	1	No	Keep
1	2	No	Keep
2	2	Yes	Skip
3	3	No	Keep
4	4	No	Keep
5	4	Yes	Skip
6	5	No	Keep

Why This Works

Step 1: Using distinct()

The distinct() function scans the list and keeps only the first occurrence of each element, removing duplicates.

Step 2: Preserving order

distinct() keeps the original order of elements while removing duplicates, unlike converting to a set which may reorder.

Step 3: Printing the result

The unique list is printed showing the list without duplicates.

Alternative Approaches

Using toSet()

kotlin

fun main() {
    val list = listOf(1, 2, 2, 3, 4, 4, 5)
    val uniqueSet = list.toSet()
    println(uniqueSet)
}

Converts list to a set which removes duplicates but does not guarantee order.

Using LinkedHashSet

kotlin

fun main() {
    val list = listOf(1, 2, 2, 3, 4, 4, 5)
    val uniqueSet = LinkedHashSet(list)
    println(uniqueSet.toList())
}

LinkedHashSet removes duplicates and preserves insertion order.

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

Time Complexity

The distinct() function scans each element once, so it runs in linear time relative to the list size.

Space Complexity

It uses extra space to store unique elements, so space grows linearly with input size.

Which Approach is Fastest?

distinct() and toSet() both run in O(n), but distinct() preserves order, which is often preferred.

Approach	Time	Space	Best For
distinct()	O(n)	O(n)	Removing duplicates while preserving order
toSet()	O(n)	O(n)	Removing duplicates when order does not matter
LinkedHashSet	O(n)	O(n)	Removing duplicates with order preservation and set behavior