125. Valid Palindrome.js

var isPalindrome = function(s) {
    let temp = "";
    for (let x = 0; x < s.length; x++) {
        if (((s[x].charCodeAt(0) > 47 && s[x].charCodeAt(0) < 58) || 
            (s[x].charCodeAt(0) > 64 && s[x].charCodeAt(0) < 91) || 
            (s[x].charCodeAt(0) > 96 && s[x].charCodeAt(0) < 123))) {
            temp += s[x].toLowerCase();
        }
    }
    return (temp == temp.split("").reverse().join(""));
};


// The overall time complexity is O(n), where n represents the number of characters in the input string s.
// 
// The process I went through here is that I knew I could save alphanumeric characters in a temporary string and 
// compare that string to its reverse, making the overall time complexity O(n). Here is the breakdown of my code:
//     1. I start my function by initializing the temporary string temp.
//     2. Then I open a for-loop to check if a character is alphanumeric, and if it is I save it's lowercase to 
//        the temp string.
//     3. Then I return the boolean which represents the equality of comparing the temp string to its reverse.