How do you solve Valid Palindrome II on LeetCode?

Valid Palindrome II (LeetCode #680) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using two pointers is efficient for palindrome checks.

What is the brute force solution for Valid Palindrome II?

The brute force approach for Valid Palindrome II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible strings that can be formed by removing one character from the original string. If any of these strings is a palindrome, we return true. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Valid Palindrome II?

Valid Palindrome II uses the following concepts: Two Pointers, String, Greedy. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Valid Palindrome II?

Explain your thought process clearly as you code. Consider edge cases, such as strings with one character or all identical characters. Practice identifying patterns in problems to improve your problem-solving speed.

What are common mistakes when solving Valid Palindrome II?

Not considering both character skip options when a mismatch is found. Assuming a single character removal guarantees a palindrome.

#680

Valid Palindrome II

Easy

Two PointersStringGreedyTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Using a two-pointer technique, we can check for palindromic properties while allowing for one character to be skipped. This is more efficient as it avoids creating multiple substrings.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, left at the start and right at the end of the string.
2Step 2: While left is less than right, check if the characters at these pointers are equal.
3Step 3: If they are not equal, check the two possibilities: skip the left character or skip the right character.
4Step 4: If either possibility results in a palindrome, return true. If the loop completes, return true.

solution.py18 lines

1# Full working Python code
2
3def valid_palindrome(s):
4    def is_palindrome_range(left, right):
5        while left < right:
6            if s[left] != s[right]:
7                return False
8            left += 1
9            right -= 1
10        return True
11
12    left, right = 0, len(s) - 1
13    while left < right:
14        if s[left] != s[right]:
15            return is_palindrome_range(left + 1, right) or is_palindrome_range(left, right - 1)
16        left += 1
17        right -= 1
18    return True

ℹ

Complexity note: The optimal solution runs in O(n) time because we only traverse the string once with two pointers, and the space complexity is O(1) since we are not using any additional data structures.

1Using two pointers is efficient for palindrome checks.
2Skipping characters can help in validating near-palindromes.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.