Valid Palindrome LeetCode Solution

Problem – 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 LeetCode Solution in Java

public class Solution {
    public boolean isPalindrome(String s) {
        if (s.isEmpty()) {
        	return true;
        }
        int head = 0, tail = s.length() - 1;
        char cHead, cTail;
        while(head <= tail) {
        	cHead = s.charAt(head);
        	cTail = s.charAt(tail);
        	if (!Character.isLetterOrDigit(cHead)) {
        		head++;
        	} else if(!Character.isLetterOrDigit(cTail)) {
        		tail--;
        	} else {
        		if (Character.toLowerCase(cHead) != Character.toLowerCase(cTail)) {
        			return false;
        		}
        		head++;
        		tail--;
        	}
        }
        
        return true;
    }
}

Valid Palindrome LeetCode Solution in Python

def isPalindrome(self, s):
    l, r = 0, 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

Valid Palindrome LeetCode Solution in C++

bool isPalindrome(string s) {
    for (int i = 0, j = s.size() - 1; i < j; i++, j--) { // Move 2 pointers from each end until they collide
        while (isalnum(s[i]) == false && i < j) i++; // Increment left pointer if not alphanumeric
        while (isalnum(s[j]) == false && i < j) j--; // Decrement right pointer if no alphanumeric
        if (toupper(s[i]) != toupper(s[j])) return false; // Exit and return error if not match
    }
    
    return true;
}
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