Check if a String is a Palindrome in JavaScript

Question

The requirements for this task are that the code is returns a 'true' or 'false' for an input string. The string can be a simply word or a phrase. The other question does not address these needs. Please reopen and answer here. I am working on a function to check if a given string is a palindrome. My code seems to work for simple one-word palindromes but not for palindromes that feature capitalization or spaces.

function palindrome(str) 
{
    var palin = str.split("").reverse().join("");

    if (palin === str){
        return true;
    } else {
        return false;
    }
}   

palindrome("eye");//Succeeds
palindrome("Race car");//Fails

Answer 1

First the string is converted to lowercase. Also, the characters that are not the alphabet are removed. So the string comparison becomes a array, then invert it, and convert it to string again.

Step 1: str1.toLowerCase().replace(...) => "Race car" => "race car" => "racecar"
Step 2: str2.split("") => ["r","a","c","e","c","a","r"] => .reverse().join() => "racecar"
Result: str1 === str2

function palindrome(str) {
   str = str.toLowerCase().replace(/[^a-z]+/g,"");
   return str === str.split("").reverse().join("")
}

alert(palindrome("eye")); //true
alert(palindrome("Race car")); //true
alert(palindrome("Madam, I'm Adam")); //true

Answer 2

Something like if (word === word.split('').reverse().join('')) {/*its a palindrome!*/} I'd say

An isPalindrome String extention:

String.prototype.isPalindrome = function () {
  var cleaned = this.toLowerCase().match(/[a-z]/gi).reverse();
  return cleaned.join('') === cleaned.reverse().join('');
}

var result = document.querySelector('#result');
result.textContent = "'eye'.isPalindrome() => " + 'eye'.isPalindrome() +
                     "\n'Something'.isPalindrome() => " + 'Something'.isPalindrome() +
                     "\n'Race Car'.isPalindrome() => " + 'Race Car'.isPalindrome() +
                     "\n'Are we not drawn onward, we few, drawn onward to new era?'.isPalindrome() => " + 
                        'Are we not drawn onward, we few, drawn onward to new era?'.isPalindrome() +
                     "\n'Never even or odd'.isPalindrome() => " + 'Never even or odd'.isPalindrome() +
                     "\n'Never odd or even'.isPalindrome() => " + 'Never odd or even'.isPalindrome();
;

<pre id="result"></pre>

Answer 3

function palindrome(str) {
    let letters = str.split('').filter(function (str) {
        return /\S/.test(str);
    });
    let reversedLetters = str.split('').reverse().filter(function (str) {
        return /\S/.test(str);
    });

    for (let i = 0; i < letters.length; i++) {
        if (letters[i].toLowerCase() !== reversedLetters[i].toLowerCase()) {
            return false;
        }
    }
    return true;
}
console.log(palindrome("eye")); //true
console.log(palindrome('Race car')); //true

Answer 4

const palindromes = arrayOfWords.filter((item) => {
    return item === item.split('').reverse().join('');
})

This is an example :-)

Answer 5

function palindrome(str) {

  var st='';    
  st=str.replace(/[^a-z0-9]/ig,"").toLowerCase();    
  var arr=[];    
  arr=st.split('');    
  arr=arr.reverse();    
  var strr='';    
  strr=arr.join('');
  if(strr==st) {
     return true;
 }
 return false;
}    
palindrome("A man, a plan, a canal. Panama");//calling the function

Answer 6

//1. change the string to an array //2. use the reverse method //3. return the array as a string //4. return input= new reversed string

var lowerCasedString = inputString.toLowerCase(); var reversedString = lowerCasedString.split("").reverse().join(""); return reversedString === lowerCasedString;
}hope this would be helpful:

Answer 7

Palindrome using ES6

const checkPalindrome=(str)=> {
  return str.toLowerCase().trim() === str.toLowerCase().trim().split('').reverse().join('');
}
console.log(checkPalindrome("Level "))

Answer 8

As easy as possible!

function checkStrForPalindorme(string) {
  const str = string.toLowerCase().trim();
  const reversedStr = str.split("").reverse().join("");

  return str === reversedStr;
}

console.log(checkStrForPalindorme("aaffaa "))
console.log(checkStrForPalindorme("dummy"))

Answer 9

I have created a lookup object first and then analyze the frequency of each lookup item

if the str = "PEEPs", then lookUp = {p:2, e:2, s:1}

Please refer to the following function for more details

function checkPlaindrome(str) {
  const lookUp = {};
  // Create the lookUp object
  for (let char of str.toLowerCase().trim()) {
    if (char !== " ") lookUp[char] = ++lookUp[char] || 1;
  }
  // Analyse the lookup object
  const singleCharCheck = Object.values(lookUp).filter((val) => val === 1);
  const findalSingleCharCheck =
    singleCharCheck.length === 0 || singleCharCheck.length === 1;
  const noneSingleCharCheck =
    Object.values(lookUp)
      .filter((val) => val !== 1)
      .filter((val) => val % 2 !== 0).length === 0;

  return findalSingleCharCheck && noneSingleCharCheck;
}

const r = checkPlaindrome("w wPEEPs"); // true
const r1 = checkPlaindrome("w PEEPs"); // false