Introduction

The Valid Palindrome II problem involves checking whether a string can become a palindrome after deleting at most one character.

Given a string s, the task is to determine:

  • whether the string can become palindrome
  • after removing at most one character

This problem helps in understanding:

  • Two Pointer Technique
  • palindrome validation
  • conditional skipping
  • string comparison logic

Example

Input:s = "abca"
Output:
True

Explanation:
Removing 'c' gives:
"aba"
which is a palindrome.

Input:
s = "abc"

Output:
False

Explanation:
Removing one character still
cannot make the string palindrome.

Constraints

 1 <= s.length <= 10^5s contains lowercase English letters

Approach 1 : Brute Force

Explanation

The simplest way to solve this problem is:

  1. Remove each character one by one
  2. Check whether resulting string becomes palindrome

This approach works but repeated palindrome checking increases time complexity.

Steps

  1. Traverse every index.
  2. Remove current character.
  3. Check whether modified string is palindrome.
  4. If palindrome found:
    • return true
  5. Otherwise:
    • return false

Dry Run

Input:s = "abca"

Remove a:
"bca"
Not palindrome

Remove b:
"aca"
Palindrome found

Final Result:
True

Brute Force Code

Complexity Analysis

Time Complexity: O(n²)Explanation:
Palindrome checking is repeated
for every character.

Space Complexity: O(n)
Explanation:
Extra strings are created.

Approach 2 : Optimized Solution (Two Pointer Technique)

Explanation

The optimized solution uses the Two Pointer Technique.

The idea is:

  1. Compare characters from both ends
  2. If mismatch occurs:
    • skip left character
      OR
    • skip right character
  3. Check whether remaining substring becomes palindrome

This avoids repeated full traversal.

Steps

  1. Initialize:
    • left = 0
    • right = n - 1
  2. Compare characters.
  3. If mismatch occurs:
    • check skipping left
    • check skipping right
  4. Return result accordingly.

Dry Run

Input:s = "abca"

Initially:
left = 0 → a
right = 3 → a
Characters match
Move pointers
left = 1 → b
right = 2 → c
Mismatch found

Check:
Skip b → "ca"
Not palindrome
Skip c → "ba"
Not palindrome
Check remaining substring:
"aba"
Palindrome found

Final Result:
True

Optimized Code

Complexity Analysis

Time Complexity: O(n)Explanation:
String is traversed at most twice.

Space Complexity: O(1)
Explanation:
No extra data structures are used.

Edge Cases

  1. String already palindrome
  2. String contains one character
  3. String cannot become palindrome
  4. String requires deleting first character
  5. String requires deleting last character

Why This Problem is Important

This problem helps in understanding:

  1. Two Pointer Technique
  2. Conditional skipping
  3. Palindrome validation
  4. Efficient traversal
  5. String comparison logic

It is one of the most important string two-pointer interview problems.

Real-World Applications

Palindrome validation concepts are used in:

  1. Text validation systems
  2. Data verification
  3. Pattern matching
  4. DNA sequence analysis
  5. String processing engines

Common Mistakes

  1. Incorrect pointer updates
  2. Forgetting skip conditions
  3. Wrong substring validation
  4. Checking more than one deletion

Interview Tips

Interviewers often expect:

  1. Two Pointer optimization
  2. O(n) solution explanation
  3. Proper mismatch handling

Always explain why only one mismatch handling is required.

Related Questions

  1. Palindrome String
  2. Reverse String
  3. Reverse Vowels
  4. Merge Strings Alternately
  5. Valid Palindrome

Final Takeaway

The Valid Palindrome II problem is a fundamental two-pointer string problem that teaches conditional skipping and efficient palindrome validation techniques. Understanding this problem builds a strong foundation for advanced string comparison and traversal interview problems.