We use recursion with two pointers (left and right) to swap characters from both ends, moving inward until they meet.

1. Define a helper function with left and right indices.
2. Base case: If left >= right, return.
3. Swap s[left] and s[right].
4. Recurse with left + 1 and right - 1.

• Time Complexity: O(N).
• Space Complexity: O(N) (recursion stack).

def reverse_string(s):
    def helper(left, right):
        if left >= right:
            return
        s[left], s[right] = s[right], s[left]
        helper(left + 1, right - 1)
    
    helper(0, len(s) - 1)