Valid Palindrome II
Given a string s, return True if s can be a palindrome after deleting at most one character.
Example:
aba
True
- The input string
sis "aba", which is already a palindrome since it reads the same backward as forward. - No character needs to be deleted to make it a palindrome.
- The condition of deleting at most one character is satisfied, as no deletion is required.
- The function returns
True, indicating that the string can be a palindrome after deleting at most one character.
Constraints:
- 1 <= len(s) <= 10^5
- s consists of lowercase English letters
Background Knowledge
The problem "Valid Palindrome II" involves determining whether a given string can be transformed into a palindrome by deleting at most one character. A palindrome is a sequence that reads the same backward as forward. To approach this problem, it's essential to understand the concept of palindromes and how to check if a string is a palindrome. This can be done by comparing characters from the start and end of the string, moving towards the center.
In the context of this problem, we need to consider the impact of deleting a character on the overall structure of the string. This involves understanding how to efficiently compare substrings and make decisions based on the presence or absence of certain characters. The problem can be solved using a combination of string manipulation and comparison techniques.
The key concept here is to identify the conditions under which deleting a character would result in a palindrome. This requires analyzing the string for any mismatches between characters from the start and end, and determining whether removing one character can resolve these mismatches.
Algorithm/Approach
The general approach to solving this type of problem involves using a two-pointer technique. This technique is commonly used in string and array problems, where two pointers are used to traverse the data structure from both ends towards the center. In this case, the two pointers can be used to compare characters from the start and end of the string, and to identify any mismatches.
Continue the full explanation
You're reading the free preview. Unlock the complete walkthrough, the code editor, test runner and reference solution with Premium.
Editor locked
The code editor is locked for Pro problems. It is only available for free problems. Please upgrade to gain access to the code editor for all problems.