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. Stops when the indices meet or cross. Demonstrates the two-pointer pattern with the smallest possible state.

Algorithm

Canonical input [1, 2, 3, 4, 5, 6, 7] (odd length, middle element stays put) yields three swap frames and reverses to [7, 6, 5, 4, 3, 2, 1].

two pointers `left` starts at index `0`, `right` starts at `arr.length - 1`. Each loop iteration swaps `arr[left]` and `arr[right]` and moves the pointers toward each other.

Basic Implementation

basic.js
Replay: real traced execution (multi-file project)
const arr = [1, 2, 3, 4, 5, 6, 7];
let left = 0;
let right = arr.length - 1;
while (left < right) {
    const tmp = arr[left];
    arr[left] = arr[right];
    arr[right] = tmp;
    left = left + 1;
    right = right - 1;
}
console.log(JSON.stringify(arr));

  1. arr ← [1, 2, 3, 4, 5, 6, 7]

    1const arr = [1, 2, 3, 4, 5, 6, 7];2let left = 0;
    values this step[1, 2, 3, 4, 5, 6, 7]arr

  2. left ← 0

    1const arr = [1, 2, 3, 4, 5, 6, 7];2let left = 0;3let right = arr.length - 1;
    values this step0left[1, 2, 3, 4, 5, 6, 7]arr

  3. right ← 6

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

  4. arr ← [7, 2, 3, 4, 5, 6, 1]

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

  5. left ← 1

    7arr[right] = tmp;8left = left + 1;9right = right - 1;
    values this step0 → 1left

  6. right ← 5

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

  7. arr ← [7, 6, 3, 4, 5, 2, 1]

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

  8. left ← 2

    7arr[right] = tmp;8left = left + 1;9right = right - 1;
    values this step1 → 2left

  9. right ← 4

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

  10. arr ← [7, 6, 5, 4, 3, 2, 1]

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

  11. left ← 3

    7arr[right] = tmp;8left = left + 1;9right = right - 1;
    values this step2 → 3left

  12. right ← 3

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

  13. while (left < right)

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

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • JavaScript: use a temporary const tmp to swap two array slots.
  • Never call arr.reverse(); the lesson is teaching the two-pointer walk.
  • The replay shows both left and right, the values about to be swapped, and the array contents after the swap. The loop-exit frame is the moment the pointers meet.