Palindromes are a classic and intriguing concept in computer science and linguistics. The problem “Valid Palindrome II” on Leetcode presents an interesting twist on standard palindrome checking, and understanding how to tackle this problem is essential for mastering string manipulation algorithms. This article provides a detailed walkthrough of the problem, the underlying concepts required to solve it, and different algorithms for tackling the problem with Python.
Given a non-empty string
s, you may delete at most one character. Judge whether you can make it a palindrome.
Explanation: You could delete the character ‘c’.
- The string will only contain lowercase characters.
- The maximum length of the string is 50,000.
A palindrome is a word, number, or other sequences of characters that reads the same forward and backward (ignoring spaces, punctuation, and capitalization).
2. Two Pointers Technique
The two pointers technique is an approach where two pointers iterate through an array (or a string) from the beginning and end, moving towards each other. This technique is especially useful when dealing with array or string manipulation problems.
1. Two Pointers with Recursion
In this approach, we will use the two pointers technique to check if the string can be a palindrome. The twist in this problem, compared to standard palindrome checking, is that we can remove at most one character.
- Initialize two pointers,
right, at the beginning and end of the string, respectively.
- Compare the characters at
- If they are equal, move
leftone step forward and
rightone step backward.
- If they are not equal, check two cases: a. Skip the character at the
leftpointer and see if the rest of the string is a palindrome. b. Skip the character at the
rightpointer and see if the rest of the string is a palindrome.
- If one of the cases in step 4 is a palindrome, the entire string can become a palindrome by removing a character. Otherwise, it cannot.
Trueif it can be a palindrome, otherwise return
def validPalindrome(s): def is_palindrome_range(i, j): return all(s[k] == s[j-k+i] for k in range(i, j)) left, right = 0, len(s) - 1 while left < right: if s[left] != s[right]: return is_palindrome_range(left+1, right) or is_palindrome_range(left, right-1) left, right = left + 1, right - 1 return True
- O(n), where n is the length of the input string.
- O(1), as we are using a constant amount of extra space.
2. Dynamic Programming Approach (Optional, Not Optimal for this Problem)
Though dynamic programming is not the most optimal solution for this problem, understanding its application can be informative for solving more complex problems.
- Create a 2D boolean array
Trueif the substring
s[i:j+1]can be a palindrome.
- Initialize all
- Iterate through the string and fill in the
- Use the
dparray to check if the string can be a palindrome by removing at most one character.
def validPalindrome(s): n = len(s) dp = [[False] * n for _ in range(n)] for i in range(n): dp[i][i] = True deleted = False for length in range(2, n+1): for i in range(n-length+1): j = i + length - 1 if s[i] == s[j]: dp[i][j] = dp[i+1][j-1] or length == 2 else: deleted = True return dp[n-1] or (dp[n-2] and deleted) or (dp[n-1] and deleted)
- O(n^2), where n is the length of the input string.
- O(n^2), as we are using an additional 2D array to store the states.
The “Valid Palindrome II” problem is a classic string manipulation problem with an interesting twist. It requires understanding the concept of palindromes and employing the two pointers technique effectively. The most optimal approach involves using two pointers with recursion to check for the possibility of forming a palindrome by removing at most one character.