Two pointers from both ends, swap until they meet.
Clarify that the input is a mutable array of characters and that in-place means modifying the array directly without creating a new one. Use the two-pointer technique: one pointer at the start and one at the end, swap characters and move pointers inward until they meet. This achieves O(n) time and O(1) extra space.
Pro tip: Mention that in languages like Java, strings are immutable, so you must use a char array; also, discuss how this approach can be extended to reverse words in a string.
Confirm that the input is a mutable array of characters, that in-place modification is required, and that O(1) extra memory means no additional data structures proportional to input size.
Describe initializing two pointers: left at index 0 and right at index n-1. While left < right, swap the characters at these indices and move left forward and right backward.
Trace the algorithm on a small example like ['h','e','l','l','o'] to demonstrate how swaps occur and pointers converge, ensuring clarity.
State that the time complexity is O(n) because each character is visited once, and space complexity is O(1) since only a constant number of variables are used.
Mention handling empty arrays, single-character arrays, and even-length arrays. Optionally, discuss how to reverse words in a string using a similar technique.
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