Reverse Array In Place (Two Pointers)

Walk two indices toward each other from the ends of the array, swapping at each step. Stops when the indices meet or cross. Demonstrates the two-pointer pattern with the smallest possible state.

Algorithm

Canonical input [1, 2, 3, 4, 5, 6, 7] (odd length, middle element stays put) yields three swap frames and reverses to [7, 6, 5, 4, 3, 2, 1].

Basic Implementation

basic.kt

fun main() {
	val arr = intArrayOf(1, 2, 3, 4, 5, 6, 7)
	var left = 0
	var right = arr.size - 1
	while (left < right) {
		val tmp = arr[left]
		arr[left] = arr[right]
		arr[right] = tmp
		left = left + 1
		right = right - 1
	}
	println(arr.joinToString(prefix = "[", postfix = "]"))
}

Complexity

Time: O(n)
Space: O(1)

Implementation notes

Kotlin: explicit three-line val tmp = arr[left]; arr[left] = arr[right]; arr[right] = tmp swap keeps the move visible. The stdlib arr.reverse() would hide the lesson.
var left = 0 and var right = arr.size - 1 use plain Int indices; the left < right guard handles the meet-in-the-middle exit honestly for the odd-length canonical input.
The replay shows both left and right, the values about to be swapped, and the array contents after the swap. The loop-exit frame is the moment the pointers meet.

two pointers `left` starts at index `0`, `right` starts at `n - 1`. Each loop iteration swaps `arr[left]` and `arr[right]` and moves the pointers toward each other.