Valid Palindrome IV - Solution & Explanation

Problem Overview: You are given a string s. The task is to determine whether it can become a palindrome after changing at most two characters. A palindrome reads the same forward and backward, so the key question is how many mirrored character pairs disagree.

Approach 1: Brute Force Character Replacement (Exponential)

A straightforward idea is to try all possible ways to modify up to two characters and check whether any resulting string becomes a palindrome. For each modified candidate, run a standard palindrome check using two pointers. While conceptually simple, the number of combinations grows quickly because each position can be changed to multiple characters. This approach has exponential branching and is impractical for longer strings. Time complexity is roughly O(26^2 * n) in simplified form, with O(n) space if new strings are created during modification.

Approach 2: Two Pointers Mismatch Count (O(n))

The optimal observation: a palindrome only cares about mirrored pairs. If s[i] and s[j] differ, you can fix that pair with one character change. Start two pointers at the beginning and end of the string and move inward. Every time the characters differ, increment a mismatch counter. If the counter exceeds two, forming a palindrome within two changes becomes impossible.

This works because each mismatch pair requires exactly one modification. For example, if s[i] != s[j], changing either character resolves that pair. Since the limit is two operations, the string can tolerate at most two mismatched mirrored pairs. The algorithm simply iterates once across half the string while comparing characters from both ends.

This implementation uses the two pointers pattern combined with simple string comparisons. The pointers move inward until they meet in the middle or the mismatch count exceeds two. Time complexity is O(n) because each character pair is examined once, and space complexity is O(1) since only a few counters and indices are used.

Recommended for interviews: Interviewers expect the two-pointer mismatch counting solution. It demonstrates recognition of palindrome symmetry and the ability to reduce the problem to pair comparisons. Mentioning the brute-force idea shows understanding of the problem constraints, but implementing the O(n) two-pointer scan proves algorithmic efficiency and familiarity with common string patterns.