Easy: Check if a string is a palindrome after removing one character

Start by clarifying the exact contract for Palindrome After One Removal: input representation, mutation rules, duplicate behavior, and what should happen on invalid or empty input. The core idea is two-pointer mismatch branching proof. Work a tiny example aloud and name the invariant after each step, because that is the proof that the approach stays correct as the input grows.

A dependable implementation plan is to scan matching ends; on first mismatch test skip-left range; test skip-right range; accept if either range is palindrome. At-most-one and exactly-one deletion have different semantics. Keep the code aligned with that invariant instead of adding special cases that silently change the contract. Test the smallest input, a normal case, duplicates or ties where relevant, an adversarial boundary, and a case that exercises the failure path. The expected complexity is O(n) time and O(1) space. Explain both time and auxiliary memory, including recursion or helper structures; do not hide sorting, copying, or repeated scans. If the title omits critical constraints, say what you need to know before committing to a more specialized solution.