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.java
Replay: real traced execution (multi-file project)
public class Basic {
    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4, 5, 6, 7};
        int left = 0;
        int right = arr.length - 1;
        while (left < right) {
            int tmp = arr[left];
            arr[left] = arr[right];
            arr[right] = tmp;
            left = left + 1;
            right = right - 1;
        }
        System.out.println(java.util.Arrays.toString(arr));
    }
}

  1. arr ← [1, 2, 3, 4, 5, 6, 7]

    2public static void main(String[] args) {3    int[] arr = {1, 2, 3, 4, 5, 6, 7};4    int left = 0;
    values this step[1, 2, 3, 4, 5, 6, 7]arr

  2. left ← 0

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

  3. right ← 6

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

  4. arr ← [7, 2, 3, 4, 5, 6, 1]

    8arr[left] = arr[right];9arr[right] = tmp;10left = left + 1;
    values this step[1, 2, 3, 4, 5, 6, 7] → [7, 2, 3, 4, 5, 6, 1]arr0left6right

  5. left ← 1

    9arr[right] = tmp;10left = left + 1;11right = right - 1;
    values this step0 → 1left

  6. right ← 5

    10    left = left + 1;11    right = right - 1;12}
    values this step6 → 5right

  7. arr ← [7, 6, 3, 4, 5, 2, 1]

    8arr[left] = arr[right];9arr[right] = tmp;10left = left + 1;
    values this step[7, 2, 3, 4, 5, 6, 1] → [7, 6, 3, 4, 5, 2, 1]arr1left5right

  8. left ← 2

    9arr[right] = tmp;10left = left + 1;11right = right - 1;
    values this step1 → 2left

  9. right ← 4

    10    left = left + 1;11    right = right - 1;12}
    values this step5 → 4right

  10. arr ← [7, 6, 5, 4, 3, 2, 1]

    8arr[left] = arr[right];9arr[right] = tmp;10left = left + 1;
    values this step[7, 6, 3, 4, 5, 2, 1] → [7, 6, 5, 4, 3, 2, 1]arr2left4right

  11. left ← 3

    9arr[right] = tmp;10left = left + 1;11right = right - 1;
    values this step2 → 3left

  12. right ← 3

    10    left = left + 1;11    right = right - 1;12}
    values this step4 → 3right

  13. while (left < right)

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

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • Java: use a temporary int tmp to swap two array slots.
  • Never call Collections.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.