Valid Palindrome LeetCode Solution | Easy Approach

Valid Palindrome A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.

Given a string s, return true if it is a palindrome, or false otherwise.

Example 1:

Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.

Example 2:

Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.

Example 3:

Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.

Constraints:

• 1 <= s.length <= 2 * 105
• s consists only of printable ASCII characters.

Valid Palindrome Solutions

✅Time: O(n)
✅Space: O(1)

C++

class Solution {
 public:
  bool isPalindrome(string s) {
    int l = 0;
    int r = s.length() - 1;

    while (l < r) {
      while (l < r && !isalnum(s[l]))
        ++l;
      while (l < r && !isalnum(s[r]))
        --r;
      if (tolower(s[l]) != tolower(s[r]))
        return false;
      ++l;
      --r;
    }

    return true;
  }
};

Java

 class Solution {
  public boolean isPalindrome(String s) {
    int l = 0;
    int r = s.length() - 1;

    while (l < r) {
      while (l < r && !Character.isLetterOrDigit(s.charAt(l)))
        ++l;
      while (l < r && !Character.isLetterOrDigit(s.charAt(r)))
        --r;
      if (Character.toLowerCase(s.charAt(l)) != Character.toLowerCase(s.charAt(r)))
        return false;
      ++l;
      --r;
    }

    return true;
  }
}

Python

class Solution:
  def isPalindrome(self, s: str) -> bool:
    l = 0
    r = len(s) - 1

    while l < r:
      while l < r and not s[l].isalnum():
        l += 1
      while l < r and not s[r].isalnum():
        r -= 1
      if s[l].lower() != s[r].lower():
        return False
      l += 1
      r -= 1

    return True

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