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.rs

fn main() {
	let arr = [1, 3, 5, 7, 9, 11, 13];
	let target = 11;
	let mut lo: i32 = 0;
	let mut hi: i32 = (arr.len() as i32) - 1;
	let mut result: i32 = -1;
	while lo <= hi {
		let mid = lo + (hi - lo) / 2;
		if arr[mid as usize] == target {
			result = mid;
			break;
		}
		if arr[mid as usize] < target {
			lo = mid + 1;
		} else {
			hi = mid - 1;
		}
	}
	println!("{}", result);
}

Complexity

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

Implementation notes

Rust: compute let mid = lo + (hi - lo) / 2;, not (lo + hi) / 2. Keeps the lesson aligned with overflow-safe practice even on small inputs.
The standard slice::binary_search would hide the window-halving loop; the lesson is teaching the loop directly. lo and hi are kept as i32 so the -1 initialisation of result is honest without Option<usize>.
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`.