In-Place Array Reversal: Why Swapping Every Index With Its Mirror Fails

A teammate wants to reverse an array in place. Their code walks over every index `i` from `0` to `n - 1` and swaps `a[i]` with its mirror element `a[n - 1 - i]`: ```python def reverse_in_place(a): n = len(a) for i in range(n): a[i], a[n - 1 - i] = a[n - 1 - i], a[i] ``` Does this reverse the array correctly? If not, explain what it actually does, fix it, and give the time and space complexity of the fixed version. This was asked as a verbal question, without running any code. ```hint Trace a small case Run the loop by hand on `[1, 2, 3, 4]` and follow what happens to the pair at positions 0 and 3 over the whole loop. ``` ### Clarifying Questions - Does "in place" mean $O(1)$ extra space? - Can the array be empty, and can its length be odd? ### What a Strong Answer Covers - A correct verdict with a precise explanation of why, based on tracing the loop - What the loop actually leaves behind - A fix with the right loop bound, including the middle element of an odd-length array - Time and space complexity of the fix, and the observation that complexity alone does not reveal the bug - Edge cases: empty, single-element and odd-length arrays ### Follow-up Questions - Rewrite the fix with two pointers moving toward each other. Should the loop condition use `<` or `<=`, and does it matter? - How would you reverse only the elements between indices `l` and `r`? - How would you rotate an array to the right by `k` positions in place, with $O(1)$ extra space?