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 `n - 1`. Each loop iteration swaps `arr[left]` and `arr[right]` and moves the pointers toward each other.

Basic Implementation

basic.cs
Replay: real traced execution (multi-file project)
using System;

class Program {
	static void Main() {
		int[] arr = new int[] { 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;
		}
		Console.WriteLine("[" + string.Join(", ", arr) + "]");
	}
}

  1. arr ← [1, 2, 3, 4, 5, 6, 7]

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

  2. left ← 0

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

  3. right ← 6

    6int left = 0;7int right = arr.Length - 1;8while (left < right) {
    values this step6right0left

  4. arr ← [7, 2, 3, 4, 5, 6, 1]

    9int tmp = arr[left];10arr[left] = arr[right];11arr[right] = tmp;
    values this step[1, 2, 3, 4, 5, 6, 7] → [7, 2, 3, 4, 5, 6, 1]arr0left6right

  5. left ← 1

    11arr[right] = tmp;12left = left + 1;13right = right - 1;
    values this step0 → 1left

  6. right ← 5

    12	left = left + 1;13	right = right - 1;14}
    values this step6 → 5right

  7. arr ← [7, 6, 3, 4, 5, 2, 1]

    9int tmp = arr[left];10arr[left] = arr[right];11arr[right] = tmp;
    values this step[7, 2, 3, 4, 5, 6, 1] → [7, 6, 3, 4, 5, 2, 1]arr1left5right

  8. left ← 2

    11arr[right] = tmp;12left = left + 1;13right = right - 1;
    values this step1 → 2left

  9. right ← 4

    12	left = left + 1;13	right = right - 1;14}
    values this step5 → 4right

  10. arr ← [7, 6, 5, 4, 3, 2, 1]

    9int tmp = arr[left];10arr[left] = arr[right];11arr[right] = tmp;
    values this step[7, 6, 3, 4, 5, 2, 1] → [7, 6, 5, 4, 3, 2, 1]arr2left4right

  11. left ← 3

    11arr[right] = tmp;12left = left + 1;13right = right - 1;
    values this step2 → 3left

  12. right ← 3

    12	left = left + 1;13	right = right - 1;14}
    values this step4 → 3right

  13. while (left < right)

    7int right = arr.Length - 1;8while (left < right) {9	int tmp = arr[left];
    values this step[7, 6, 5, 4, 3, 2, 1]arr3left3right

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • C#: explicit three-line int tmp = arr[left]; arr[left] = arr[right]; arr[right] = tmp; swap keeps the move visible. Array.Reverse(arr) would hide the lesson.
  • int left = 0; and int right = arr.Length - 1; use plain int indices against a managed reference array; each indexed read/write is bounds-checked by the CLR. The left < right guard handles the meet-in-the-middle exit honestly for the odd-length canonical input.
  • 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.