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.java

public class Basic {
    public static void main(String[] args) {
        int[] arr = {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;
            }
        }
        System.out.println(result);
    }
}

Complexity

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

Implementation notes

Java: compute mid = lo + (hi - lo) / 2, not (lo + hi) / 2. Keeps the lesson aligned with overflow-safe practice even on small inputs.
Never call Arrays.binarySearch(...); the lesson is teaching the window-halving loop.
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`.