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With two pointers, we've cut the number of operations in half. It's much faster now! However, similar to the brute force, the time complexity is still O(n).

Why Is This?

Why Is This?

Well, if n is the length of the string, we'll end up making n/2 swaps. But remember, Big O Notation isn't about the raw number of operations required for an algorithm-- it's about how the number scales with the input.

So despite requiring half the number operations-- a 4-character string would require 2 swaps with the two-pointer method. But an 8-character string would require 4 swaps. The input doubled, and so did the number of operations.

If you haven't by now, try to do the problem before moving on.