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.cs

using System;

class Program {
	static void Main() {
		int[] arr = new int[] { 1, 3, 5, 7, 9, 11, 13 };
		int target = 11;
		int lo = 0;
		int hi = arr.Length - 1;
		int result = -1;
		while (lo <= hi) {
			int mid = lo + (hi - lo) / 2;
			if (arr[mid] == target) {
				result = mid;
				break;
			}
			if (arr[mid] < target) {
				lo = mid + 1;
			} else {
				hi = mid - 1;
			}
		}
		Console.WriteLine(result);
	}
}

Complexity

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

Implementation notes

C#: compute int mid = lo + (hi - lo) / 2;, not (lo + hi) / 2. Keeps the lesson aligned with overflow-safe practice even on small inputs.
Array.BinarySearch(arr, 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`.