Palindromes are among the most interesting problems in algorithmic challenges, and one variant of it that often comes up in interviews and coding competitions is the Valid Palindrome II problem. In this blog, we’ll break down the problem, explore different approaches to solve it, and walk through the code implementation in Python.

Problem Statement: Valid Palindrome II

Given a string s, determine if it can become a palindrome after deleting at most one character. A string is a palindrome if it reads the same forward and backward.

Example 1:

Input: "abca"
Output: True
Explanation: After deleting the character "c", the string becomes "aba", which is a palindrome.

Example 2:

Input: "racecar"
Output: True
Explanation: The input is already a palindrome.

Example 3:

Input: "hello"
Output: False
Explanation: Removing one character will not make this a palindrome.

Key Insights

In the Valid Palindrome II problem, we can either have the string be a palindrome already, or we can remove one character and check if it can become one. This problem is a variation of the standard palindrome check, which is done by comparing characters from the start and end of the string, moving inward.

Approach to Solve the Problem

1. Two-Pointer Technique

The two-pointer technique is a common strategy for solving palindrome problems. Here’s how it works:

  • Start by placing two pointers: one at the beginning (left) and one at the end (right) of the string.
  • Compare the characters at both pointers.
  • If they match, move both pointers inward (i.e., increment left and decrement right).
  • If they don’t match, try to skip either the left character or the right character, and check if the resulting substring is a palindrome.

This is the key insight: if we encounter a mismatch, we have two options for removal—either the left character or the right character. If removing one of these results in a palindrome, then the answer is True.

Algorithm Steps

  1. Start with two pointers: left at the beginning and right at the end of the string.
  2. If the characters at left and right are the same, move inward (increment left and decrement right).
  3. If the characters don’t match, check if the substring formed by skipping either the left character (s[left+1:right+1]) or the right character (s[left:right]) is a palindrome.
  4. Return True if removing one character makes the string a palindrome, otherwise return False.

Code Implementation

Here’s how we can implement this solution in Python:

def validPalindrome(s):
    # Helper function to check if a substring is a palindrome
    def is_palindrome(sub):
        return sub == sub[::-1]

    left, right = 0, len(s) - 1

    while left < right:
        if s[left] == s[right]:
            left += 1
            right -= 1
        else:
            # Check by skipping one character from either end
            return is_palindrome(s[left + 1:right + 1]) or is_palindrome(s[left:right])

    return True

# Example usage:
print(validPalindrome("abca"))  # Output: True
print(validPalindrome("racecar"))  # Output: True
print(validPalindrome("hello"))  # Output: False

Explanation of the Code

  1. Helper Function: We define a helper function is_palindrome() that checks if a given string or substring is a palindrome by comparing it to its reverse (sub == sub[::-1]).
  2. Two Pointers:
  • We initialize two pointers, left at the start and right at the end of the string.
  • We move these pointers inward as long as the characters at both pointers are equal.
  1. Mismatch Handling:
  • If we encounter a mismatch (i.e., s[left] != s[right]), we check whether skipping the left character (s[left + 1:right + 1]) or the right character (s[left:right]) results in a palindrome.
  • If either substring is a palindrome, the function returns True.
  1. Final Check: If no mismatches are found, or if skipping one character makes the string a palindrome, the function returns True.

Example Walkthrough

Let’s walk through the code using the input s = "abca":

  1. Initial State: left = 0, right = 3
  • s[left] = "a", s[right] = "a" → Characters match, so move inward.
  • left = 1, right = 2
  1. Second State: s[left] = "b", s[right] = "c"
  • Characters do not match. We now check two cases:
    • Is the substring s[left + 1:right + 1] = "ba" a palindrome? No.
    • Is the substring s[left:right] = "bc" a palindrome? Yes.
  • Since one of the substrings is a palindrome, the function returns True.

Time Complexity Analysis

The time complexity of the solution is (O(n)), where n is the length of the string. Here’s why:

  • We traverse the string once using the two-pointer technique.
  • The helper function is_palindrome() operates on a substring, and in the worst case, we check two substrings of length (O(n)).
  • Therefore, the overall complexity is linear, (O(n)).

The space complexity is (O(1)) since we only use a few pointers and don’t allocate extra space for data structures (other than the substrings temporarily created during the mismatch check).

Edge Cases

  1. Empty String: An empty string is trivially a palindrome, so the result is True.
  2. Single Character String: A single character string is also always a palindrome.
  3. String Without Mismatch: If the string is already a palindrome, the function will return True without needing to remove any character.
  4. Multiple Mismatches: If more than one mismatch is found, the function will correctly return False, as removing just one character won’t be enough to make the string a palindrome.

The Valid Palindrome II problem offers an interesting twist on the classic palindrome check by allowing for the removal of one character. Using the two-pointer technique and a helper function to check substrings, we can efficiently solve the problem in linear time. This approach handles typical cases as well as edge cases, making it a robust solution for interview or competitive coding settings.

By understanding the structure and logic behind the two-pointer method, you’ll be better equipped to handle similar problems that require string manipulation and palindrome checks.

Frequently Asked Questions (FAQs)

Q1: What is a palindrome?
A palindrome is a string that reads the same forward and backward. For example, “racecar” is a palindrome because it is identical when reversed.

Q2: What does “Valid Palindrome II” mean?
It refers to a problem where you are asked to check whether a string can be converted into a palindrome by removing at most one character.

Q3: Can we solve the problem using recursion instead of two pointers?
Yes, but recursion would likely result in a less efficient solution. The two-pointer method is more optimal in terms of time and space complexity.

Q4: What if more than one character needs to be removed to make the string a palindrome?
The problem specifically asks if the string can be a palindrome after removing at most one character. If more than one character needs to be removed, the answer is False.

Q5: What is the time complexity of this solution?
The time complexity of this solution is (O(n)), where n is the length of the string.

By mastering the logic and approach to this problem, you’ll be well-equipped to tackle more complex string problems in competitive programming and interviews.

Read More …

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