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The two-pointer swap is pretty standard but I spent an embarrassing amount of time second-guessing whether swapping with a temp variable counted as 'extra space.' It doesn't, obviously.
Use a two-pointer technique: initialize left at 0 and right at len(lst)-1, then swap elements while left < right, incrementing left and decrementing right. This achieves O(n) time and O(1) extra space. Then demonstrate with empty, single-element, and six-element lists, and verify in-place mutation by checking the list's id before and after.
Pro tip: Explicitly state that the function returns None to emphasize in-place modification, and mention that the id check confirms no new list object is created. Also, note that the algorithm handles edge cases naturally without special branching.
Restate the problem: reverse a list in-place, O(n) time, O(1) extra space, no slicing or built-in reverse. Confirm that the function should mutate the input list and return None.
Explain the approach: use two indices, left starting at 0 and right at len(lst)-1. While left < right, swap lst[left] and lst[right], then move left forward and right backward.
Write the Python function with a clear name like reverse_in_place. Include a docstring specifying time and space complexity. Ensure no extra data structures are used.
Test with an empty list, a single-element list, and a six-element list (e.g., [1,2,3,4,5,6]). Show that the function mutates the list correctly and returns None.
Capture the id of the list before and after calling the function, and assert they are equal. This proves the list object was not replaced.
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