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].

Basic Implementation

Basic.java

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));
    }
}

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.

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.