Searching

Binary Search (Iterative)

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

Algorithm

Basic Implementation

basic.dart 
void main() {
  final arr = <int>[1, 3, 5, 7, 9, 11, 13];
  final target = 11;
  var lo = 0;
  var hi = arr.length - 1;
  var result = -1;
  while (lo <= hi) {
    final mid = lo + (hi - lo) ~/ 2;
    if (arr[mid] == target) {
      result = mid;
      break;
    }
    if (arr[mid] < target) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  print(result);
}

The pinned run searches for 11 in [1, 3, 5, 7, 9, 11, 13]. The diagrams highlight the inclusive [lo, hi] window and each midpoint.

Step 1 - First midpoint is too small

lo = 0, hi = 6, mid = 3, and arr[3] = 7 is below target 11.

Probe 1 keeps the right half.i0i1i2i3i4i5i6135791113lomidtargethi

Step 2 - Window narrows to the right

Because 7 < 11, set lo = 4 and keep hi = 6.

After discarding indexes 0 through 3.i0i1i2i3i4i5i6135791113discarddiscarddiscarddiscardlomidhi

Step 3 - Second midpoint matches

Now mid = 5 and arr[5] = 11, so the algorithm returns index 5.

Probe 2 finds target 11 at index 5.i4i5i6return911135lomid == targethiindex

Complexity

  • Time: O(log n)
  • Space: O(1)

Implementation notes

  • Dart: use integer division (~/) for mid. Do not call binarySearch from package:collection; it hides the loop the lesson is teaching, and the goal here is zero pub dependencies.
  • The replay highlights the [lo, hi] window, mid, and which branch (left half / right half / match) the step takes.
inclusive bounds `lo` and `hi` are both inclusive; the loop runs while `lo <= hi`.
overflow-safe midpoint `mid = lo + (hi - lo) ~/ 2` avoids `(lo + hi)` overflow on large inputs.