Compared to an O(1) space palindrome check, what is the space complexity of the stack-based approach?

• A. O(1) space
• B. O(n log n) space
• C. O(n) space
• D. O(n^2) space

Answer: (C)

The main idea here is how much extra storage a stack-based palindrome check needs as the input grows. With a stack-based approach you push characters onto the stack so you can pop them in reverse order to compare with the second half of the string. For a string of length n, you end up storing about n characters (or n/2 if you stop at the middle, but that’s still proportional to n). So the extra space grows linearly with the input, giving O(n) space. In contrast, the constant-space version uses only a couple of indices or pointers and a few variables, hence O(1) space. The other options imply sublinear or superlinear growth that doesn’t match how a stack holds elements, so they aren’t correct.