




Sponsored
Sponsored
This approach employs two pointers: one starting at the beginning of the string and the other at the end. The algorithm checks if the characters at these pointers are the same, ignoring case and non-alphanumeric characters. If they match, both pointers move inward. If they don't or if any pointer surpasses the other, the string is not a palindrome.
Time Complexity: O(n), where n is the length of the string, because we go through the string at most once. 
Space Complexity: O(1), as we use a constant amount of space.
1class Solution:
2    def isPalindrome(self, s: str) -> bool:
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
14
15# Usage
16solution = Solution()
17print(solution.isPalindrome("A man, a plan, a canal: Panama"))This Python function checks if a given string is a palindrome by comparing characters from both directions. It skips non-alphanumeric characters and compares the rest while ignoring case, returning the correct result accordingly.
This approach first cleans the input string by removing non-alphanumeric characters and converting all letters to lowercase. It then compares the normalized string to its reverse to determine if it is a palindrome.
Time Complexity: O(n), where n is the length of the string. 
Space Complexity: O(n), because the cleaned string and its reverse require extra space proportional to n.
1#include <iostream>
2#include <string>
3#include <algorithm>
4#include <cctype>
bool isPalindrome(std::string s) {
    std::string filtered = "";
    for (char c : s) {
        if (std::isalnum(c)) {
           filtered += std::tolower(c);
        }
    }
    std::string reversed = filtered;
    std::reverse(reversed.begin(), reversed.end());
    return filtered == reversed;
}
int main() {
    std::string s = "A man, a plan, a canal: Panama";
    if (isPalindrome(s)) {
        std::cout << "true" << std::endl;
    } else {
        std::cout << "false" << std::endl;
    }
    return 0;
}This C++ solution constructs a cleaned version of the input, removing non-alphanumeric characters and converting them to lowercase. It checks if this cleaned string equals its reverse.