Queue Enqueue/Dequeue

Algorithm

The replay uses the same three values in every language, so this Kotlin DSA implementation can be compared directly with the rest of the DSA track.

front The front is the oldest value still waiting in the queue.

FIFO A queue removes values in first-in, first-out order.

Visual walkthrough

The queue keeps the oldest value at the front and adds new values at the back.

Step 1 - Enqueue 10, 20, 30

New values join at the back. The oldest value, 10, stays at the front.

Step 2 - Dequeue removes 10

Removing from the front returns 10 and makes 20 the new front.

Basic Implementation

                            basic.kt
                        

                            import java.util.ArrayDeque

fun render(values: List<Int>) = values.joinToString(" -> ")

fun main() {
    val queue = ArrayDeque<Int>()
    for (value in listOf(10, 20, 30)) queue.addLast(value)
    val removed = mutableListOf<Int>()
    while (!queue.isEmpty()) removed.add(queue.removeFirst())
    println(render(removed))
}

                        

Complexity

Time: O(1) per operation with a real queue
Space: O(n)

Implementation notes

Kotlin imports java.util.ArrayDeque and creates val queue = ArrayDeque<Int>(); the reference is not rebound, but the deque contents mutate.
Enqueue uses queue.addLast(value) for values from listOf(10, 20, 30), preserving insertion order at the back of the deque.
Removed values are collected in val removed = mutableListOf<Int>(), so the output list mutates separately from the queue.
Dequeue uses queue.removeFirst() inside while (!queue.isEmpty()); the emptiness guard avoids the exception path for removing from an empty deque.
Values are non-null Ints throughout; no nullable queue result is used.
The trace shows the queue moving from [] to [10, 20, 30], then removing 10 with queue [20, 30], and finally removed [10, 20, 30] with queue [].
render(values: List<Int>) uses joinToString(" -> "), and println(render(removed)) prints 10 -> 20 -> 30.