Binary Search (Iterative)

On a sorted array, narrow [lo, hi] window by halving until arr[mid] equals the target or the window is empty. Demonstrates the "discard half the search space" invariant.

Algorithm

Canonical input arr = [1, 3, 5, 7, 9, 11, 13] with target = 11 finishes in two frames: descend right to mid = 5, match.

Basic Implementation

basic.kt

fun main() {
	val arr = intArrayOf(1, 3, 5, 7, 9, 11, 13)
	val target = 11
	var lo = 0
	var hi = arr.size - 1
	var result = -1
	while (lo <= hi) {
		val mid = lo + (hi - lo) / 2
		if (arr[mid] == target) {
			result = mid
			break
		}
		if (arr[mid] < target) {
			lo = mid + 1
		} else {
			hi = mid - 1
		}
	}
	println(result)
}

Complexity

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

Implementation notes

Kotlin: compute val mid = lo + (hi - lo) / 2, not (lo + hi) / 2. Keeps the lesson aligned with overflow-safe practice even on small inputs.
arr.binarySearch(target) would hide the window-halving loop; the lesson is teaching the loop directly. lo and hi are kept as Int so the -1 initialisation of result is honest without Int?.
The replay highlights the [lo, hi] window, marks mid, and labels the branch taken (left half / right half / match).

midpoint `mid = lo + (hi - lo) / 2` (overflow-safe integer division).

inclusive window `hi` is inclusive. The loop runs while `lo <= hi`.