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.c
#include <stdio.h>

int main(void) {
    int arr[] = {1, 2, 3, 4, 5, 6, 7};
    size_t n = sizeof(arr) / sizeof(arr[0]);
    size_t left = 0;
    size_t right = n - 1;
    while (left < right) {
        int tmp = arr[left];
        arr[left] = arr[right];
        arr[right] = tmp;
        left = left + 1;
        right = right - 1;
    }
    printf("[");
    for (size_t i = 0; i < n; ++i) {
        if (i > 0) printf(", ");
        printf("%d", arr[i]);
    }
    printf("]\n");
    return 0;
}

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • C: use a temporary int tmp to swap two slots. C has no built-in reverse helper, so the lesson stays directly on the two-pointer walk.
  • size_t left and size_t right mirror sizeof's return type and the size discipline used by n; 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.
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.