Palindrome Check

Given a string, return True if it reads the same forward and backward, and False otherwise. For example, "racecar" is a palindrome, "hello" is not. A single character like "a" is always a palindrome.

A palindrome is symmetric: $s[i] = s[N-1-i]$ for all $i$. This is the same two-pointer technique used for array reversal — compare from both ends and move inward. Finding any mismatch lets us immediately return False (early exit).

1. Initialize two pointers: left = 0, right = len(s) - 1.
2. While left < right, compare s[left] with s[right].
3. If they differ, return False immediately.
4. If they match, move left forward and right backward.
5. If the loop completes without mismatches, return True.

• Comparing with s == s[::-1] — correct but uses $O(N)$ extra space
• Not handling case sensitivity: "Racecar" returns False unless normalized to lowercase first
• Forgetting edge cases: empty string "" and single character "a" are both palindromes

This problem reinforces the two-pointer pattern from Day 4 (Array Reversal). Strings in Python are essentially character arrays, so the same pointer logic applies.