How do you solve Valid Palindrome on LeetCode?

Valid Palindrome (LeetCode #125) 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: Cleaning the string is essential for accurate palindrome checking.

What is the brute force solution for Valid Palindrome?

The brute force approach for Valid Palindrome is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves cleaning the string by removing non-alphanumeric characters and converting it to lowercase, then checking if the cleaned string is the same when reversed. This is straightforward but not efficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

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

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

What are the interview tips for Valid Palindrome?

Always clarify the problem requirements and constraints before coding. Think about edge cases, such as empty strings or strings with only non-alphanumeric characters. Explain your thought process while coding to demonstrate your understanding.

What are common mistakes when solving Valid Palindrome?

Not handling case sensitivity properly. Failing to skip non-alphanumeric characters correctly.

#125

Valid Palindrome

Easy

Two PointersStringTwo 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)

The optimal approach uses two pointers to check characters from both ends of the string towards the center. This avoids the need for extra space for a reversed string and is more efficient.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, one at the start and one at the end of the string.
2Step 2: Move the pointers towards the center, skipping non-alphanumeric characters.
3Step 3: Compare the characters at both pointers. If they are not the same, return false. If they are the same, continue until the pointers meet.
4Step 4: If all characters match, return true.

solution.py13 lines

1# Full working Python code
2def isPalindrome(s):
3    left, right = 0, len(s) - 1
4    while left < right:
5        while left < right and not s[left].isalnum():
6            left += 1
7        while left < right and not s[right].isalnum():
8            right -= 1
9        if s[left].lower() != s[right].lower():
10            return False
11        left += 1
12        right -= 1
13    return True

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once with two pointers. The space complexity is O(1) since we are not using any additional data structures that grow with input size.

1Cleaning the string is essential for accurate palindrome checking.
2Using two pointers is an efficient way to compare characters without extra space.

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