Arrays and Iteration

Reverse Array In Place (Two Pointers)

Walk two indices toward each other from the ends of the array, swapping at each step. The two-pointer pattern with the smallest possible state. The loop stops when the indices meet or cross.

Algorithm

The canonical input [1, 2, 3, 4, 5, 6, 7] reverses to [7, 6, 5, 4, 3, 2, 1] after three swaps. The middle element at index 3 is untouched because the pointers meet there.

two pointers Indices walk toward each other and swap.

Basic Implementation

basic.dart
Replay: real traced execution (multi-file project)
void main() {
  final arr = <int>[1, 2, 3, 4, 5, 6, 7];
  var left = 0;
  var right = arr.length - 1;
  while (left < right) {
    final tmp = arr[left];
    arr[left] = arr[right];
    arr[right] = tmp;
    left = left + 1;
    right = right - 1;
  }
  print(arr);
}

  1. arr ← [1, 2, 3, 4, 5, 6, 7]

    1void main() {2  final arr = <int>[1, 2, 3, 4, 5, 6, 7];3  var left = 0;
    values this step[1, 2, 3, 4, 5, 6, 7]arr

  2. left ← 0

    2final arr = <int>[1, 2, 3, 4, 5, 6, 7];3var left = 0;4var right = arr.length - 1;
    values this step0left[1, 2, 3, 4, 5, 6, 7]arr

  3. right ← 6

    3var left = 0;4var right = arr.length - 1;5while (left < right) {
    values this step6right[1, 2, 3, 4, 5, 6, 7]arr0left

  4. arr ← [7, 2, 3, 4, 5, 6, 1]

    6final tmp = arr[left];7arr[left] = arr[right];8arr[right] = tmp;
    values this step[1, 2, 3, 4, 5, 6, 7] → [7, 2, 3, 4, 5, 6, 1]arr0left6right

  5. left ← 1

    8arr[right] = tmp;9left = left + 1;10right = right - 1;
    values this step0 → 1left

  6. right ← 5

    9  left = left + 1;10  right = right - 1;11}
    values this step6 → 5right

  7. arr ← [7, 6, 3, 4, 5, 2, 1]

    6final tmp = arr[left];7arr[left] = arr[right];8arr[right] = tmp;
    values this step[7, 2, 3, 4, 5, 6, 1] → [7, 6, 3, 4, 5, 2, 1]arr1left5right

  8. left ← 2

    8arr[right] = tmp;9left = left + 1;10right = right - 1;
    values this step1 → 2left

  9. right ← 4

    9  left = left + 1;10  right = right - 1;11}
    values this step5 → 4right

  10. arr ← [7, 6, 5, 4, 3, 2, 1]

    6final tmp = arr[left];7arr[left] = arr[right];8arr[right] = tmp;
    values this step[7, 6, 3, 4, 5, 2, 1] → [7, 6, 5, 4, 3, 2, 1]arr2left4right

  11. left ← 3

    8arr[right] = tmp;9left = left + 1;10right = right - 1;
    values this step2 → 3left

  12. right ← 3

    9  left = left + 1;10  right = right - 1;11}
    values this step4 → 3right

  13. while (left < right)

    4var right = arr.length - 1;5while (left < right) {6  final tmp = arr[left];
    values this step[7, 6, 5, 4, 3, 2, 1]arr3left3right

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • Dart: use the explicit tmp = arr[left]; arr[left] = arr[right]; arr[right] = tmp; triple. Avoid arr.reversed.toList(), List.from(arr.reversed), or assigning to arr from a reversed view; all hide the step-by-step pointer walk the lesson is teaching.
  • Replay highlights both left and right per frame plus the new array contents after each swap, matching the lesson spec.