680. Valid Palindrome II
You can find this on LeetCode here , solution is here, and understand the logic here!
Thoughts and Talk
Palindromes have stood the test of time as a classic interview question due to their fundamental nature. Although there are various approaches to solving them, the traditional two-pointer method remains the simplest. However, in this current referenced leetcode problem there’s a twist to this problem❗ -it’s not just about determining if a string is a palindrome. The real challenge lies in identifying whether removing one character from the string would transform it into a palindrome.
This added complexity makes the question intriguing, even if it’s categorized as easy. It requires a keen eye for detail and a thorough understanding of palindrome properties to solve effectively.
The problem of determining if a string can be transformed into a palindrome by removing at most one character introduces a fascinating twist on a classic problem. This task isn’t just about palindrome detection; it also tests our ability to handle slight modifications in string properties efficiently. It’s a problem that requires not just understanding basic string manipulation but also a deeper insight into decision-making processes in algorithms.
Problem Statement
We need to determine whether a given string can be rendered a palindrome by deleting at most one character. This problem extends the classic palindrome check by adding the possibility of minor adjustments, presenting an interesting challenge in algorithm design.
Hints
- Utilize two pointers to compare characters from the beginning and end of the string.
- Upon finding non-matching characters, decide whether skipping the current character from the beginning or the end might help.
- Only one such skip is allowed. If skipping doesn’t help, the answer should be false.
Approaches Explored
Approach 1: Two-pointer with single skip check
Pros:
- Efficient, since it typically requires scanning the string at most twice.
- Simple to implement using conditional checks to handle the single allowed skip.
Cons:
Approach 2: Modified two-pointer with state tracking (not so different!)
This is the approach reflected in our original code. ouIt uses a breaker variable to handle complex decision points when both characters to the left and right could potentially be skipped to maintain a palindrome.
Originally published at https://www.thetechcruise.com/680-valid-palindrome-ii/
