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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.
1using System;
2
3public class Solution {
4    public bool IsPalindrome(string s) {
5        int left = 0, right = s.Length - 1;
6        while (left < right) {
7            while (left < right && !Char.IsLetterOrDigit(s[left])) left++;
8            while (left < right && !Char.IsLetterOrDigit(s[right])) right--;
9            if (Char.ToLower(s[left]) != Char.ToLower(s[right])) return false;
10            left++;
11            right--;
12        }
13        return true;
14    }
15
16    public static void Main() {
17        Solution solution = new Solution();
18        string s = "A man, a plan, a canal: Panama";
19        Console.WriteLine(solution.IsPalindrome(s));
20    }
21}This C# method verifies if a string is a palindrome, maneuvering two pointers and comparing respective characters while disregarding their cases and skipping non-alphanumeric characters.
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.