Valid Plaindrome

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Unit 4 Session 2 (Click for link to problem statements)

U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• What happens if the input string is empty?
  • The function should return True, as an empty string or a string with a single character is trivially a palindrome.
• How does the function handle case sensitivity and non-alphanumeric characters?
  • Assuming standard palindrome conditions where only alphanumeric characters are considered and they are case insensitive. However, this specific implementation does not address these issues and treats the string as is.

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a helper function to check for palindrome by comparing characters from both ends. If a mismatch is found, check the possibility of forming a palindrome by removing one character either from the left or the right.

1) Start with two pointers at the beginning and end of the string.
2) While the two pointers haven't met:
  a) If characters at the pointers match, move both pointers towards the center.
  b) If they don't match:
     i) Check if skipping the character at the left pointer results in a palindrome.
    ii) Check if skipping the character at the right pointer results in a palindrome.
    iii) Return True if either of the above checks returns True.
3) Return True if no mismatches were severe enough to prevent a palindrome (after considering one removal).

⚠️ Common Mistakes

• Not considering both possibilities of removing characters when a mismatch is found.
• Mismanaging pointers which could lead to incorrect comparisons or missing the chance to verify potential palindromes.

I-mplement

def is_palindrome(s, left, right):
    while left < right:
        if s[left] != s[right]:
            return False
        left, right = left + 1, right - 1
    return True

def valid_palindrome(s):
    left, right = 0, len(s) - 1
    
    while left < right:
        # When characters mismatch, check the two possibilities
        if s[left] != s[right]:
            # Check by skipping left character or skipping right character
            return is_palindrome(s, left + 1, right) or is_palindrome(s, left, right - 1)
        left, right = left + 1, right - 1
    
    # If it's already a palindrome or can be one by removing a character
    return True