How to write palindrome in JavaScript

Question

I wonder how to write palindrome in javascript, where I input different words and program shows if word is palindrome or not. For example word noon is palindrome, while bad is not.

Thank you in advance.

Answer 1

function palindrome(str) {

    var len = str.length;
    var mid = Math.floor(len/2);

    for ( var i = 0; i < mid; i++ ) {
        if (str[i] !== str[len - 1 - i]) {
            return false;
        }
    }

    return true;
}

palindrome will return if specified word is palindrome, based on boolean value (true/false)

UPDATE:

I opened bounty on this question due to performance and I've done research and here are the results:

If we are dealing with very large amount of data like

var abc = "asdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfd";

for ( var i = 0; i < 10; i++ ) {
    abc += abc;  // making string even more larger
}

function reverse(s) { // using this method for second half of string to be embedded
    return s.split("").reverse().join("");
}

abc += reverse(abc); // adding second half string to make string true palindrome

In this example palindrome is True, just to note

Posted palindrome function gives us time from 180 to 210 Milliseconds (in current example), and the function posted below with string == string.split('').reverse().join('') method gives us 980 to 1010 Milliseconds.

Machine Details:

System: Ubuntu 13.10 OS Type: 32 Bit RAM: 2 Gb CPU: 3.4 Ghz*2 Browser: Firefox 27.0.1

Answer 2

Try this:

var isPalindrome = function (string) {
    if (string == string.split('').reverse().join('')) {
        alert(string + ' is palindrome.');
    }
    else {
        alert(string + ' is not palindrome.');
    }
}

document.getElementById('form_id').onsubmit = function() {
   isPalindrome(document.getElementById('your_input').value);
}

So this script alerts the result, is it palindrome or not. You need to change the your_id with your input id and form_id with your form id to get this work.

Demo!

Answer 3

Use something like this

function isPalindrome(s) {
    return s == s.split("").reverse().join("") ? true : false;
}

alert(isPalindrome("noon"));

alternatively the above code can be optimized as [updated after rightfold's comment]

function isPalindrome(s) {
    return s == s.split("").reverse().join("");
}

alert(isPalindrome("malayalam")); 
alert(isPalindrome("english")); 

Answer 4

Faster Way:

-Compute half the way in loop.

-Store length of the word in a variable instead of calculating every time.

EDIT: Store word length/2 in a temporary variable as not to calculate every time in the loop as pointed out by (mvw) .

function isPalindrome(word){
   var i,wLength = word.length-1,wLengthToCompare = wLength/2;

   for (i = 0; i <= wLengthToCompare ; i++) {
     if (word.charAt(i) != word.charAt(wLength-i)) {
        return false;
     }
   }
   return true;
} 

Answer 5

Let us start from the recursive definition of a palindrome:

1. The empty string '' is a palindrome
2. The string consisting of the character c, thus 'c', is a palindrome
3. If the string s is a palindrome, then the string 'c' + s + 'c' for some character c is a palindrome

This definition can be coded straight into JavaScript:

function isPalindrome(s) {
  var len = s.length;
  // definition clauses 1. and 2.
  if (len < 2) {
    return true;
  }
  // note: len >= 2
  // definition clause 3.
  if (s[0] != s[len - 1]) {
    return false;
  }
  // note: string is of form s = 'a' + t + 'a'
  // note: s.length >= 2 implies t.length >= 0
  var t = s.substr(1, len - 2);
  return isPalindrome(t);
}

Here is some additional test code for MongoDB's mongo JavaScript shell, in a web browser with debugger replace print() with console.log()

function test(s) {
  print('isPalindrome(' + s + '): ' + isPalindrome(s));
}

test('');
test('a');
test('ab');
test('aa');
test('aab');
test('aba');
test('aaa');
test('abaa');
test('neilarmstronggnortsmralien');
test('neilarmstrongxgnortsmralien');
test('neilarmstrongxsortsmralien');

I got this output:

$ mongo palindrome.js
MongoDB shell version: 2.4.8
connecting to: test
isPalindrome(): true
isPalindrome(a): true
isPalindrome(ab): false
isPalindrome(aa): true
isPalindrome(aab): false
isPalindrome(aba): true
isPalindrome(aaa): true
isPalindrome(abaa): false
isPalindrome(neilarmstronggnortsmralien): true
isPalindrome(neilarmstrongxgnortsmralien): true
isPalindrome(neilarmstrongxsortsmralien): false

An iterative solution is:

function isPalindrome(s) {
  var len = s.length;
  if (len < 2) {
    return true;
  }
  var i = 0;
  var j = len - 1;
  while (i < j) {
    if (s[i] != s[j]) {
      return false;
    }
    i += 1;
    j -= 1;
  }
  return true;
}

Answer 6

Look at this:

function isPalindrome(word){
    if(word==null || word.length==0){
        // up to you if you want true or false here, don't comment saying you 
        // would put true, I put this check here because of 
        // the following i < Math.ceil(word.length/2) && i< word.length
        return false;
    }
    var lastIndex=Math.ceil(word.length/2);
    for (var i = 0; i < lastIndex  && i< word.length; i++) {
        if (word[i] != word[word.length-1-i]) {
            return false;
        }
     }
     return true;
} 

Edit: now half operation of comparison are performed since I iterate only up to half word to compare it with the last part of the word. Faster for large data!!!

Since the string is an array of char no need to use charAt functions!!!

Reference: http://wiki.answers.com/Q/Javascript_code_for_palindrome

Answer 7

Taking a stab at this. Kind of hard to measure performance, though.

function palin(word) {
    var i = 0,
        len = word.length - 1,
        max = word.length / 2 | 0;

    while (i < max) {
        if (word.charCodeAt(i) !== word.charCodeAt(len - i)) {
            return false;
        }
        i += 1;
    }
    return true;
}

My thinking is to use charCodeAt() instead charAt() with the hope that allocating a Number instead of a String will have better perf because Strings are variable length and might be more complex to allocate. Also, only iterating halfway through (as noted by sai) because that's all that's required. Also, if the length is odd (ex: 'aba'), the middle character is always ok.

Answer 8

Best Way to check string is palindrome with more criteria like case and special characters...

function checkPalindrom(str) {
    var str = str.replace(/[^a-zA-Z0-9]+/gi, '').toLowerCase();
    return str == str.split('').reverse().join('');
}

You can test it with following words and strings and gives you more specific result.
1. bob
2. Doc, note, I dissent. A fast never prevents a fatness. I diet on cod

For strings it ignores special characters and convert string to lower case.

Answer 9

The most important thing to do when solving a Technical Test is Don't use shortcut methods -- they want to see how you think algorithmically! Not your use of methods.

Here is one that I came up with (45 minutes after I blew the test). There are a couple optimizations to make though. When writing any algorithm, its best to assume false and alter the logic if its looking to be true.

isPalindrome():

Basically, to make this run in O(N) (linear) complexity you want to have 2 iterators whose vectors point towards each other. Meaning, one iterator that starts at the beginning and one that starts at the end, each traveling inward. You could have the iterators traverse the whole array and use a condition to break/return once they meet in the middle, but it may save some work to only give each iterator a half-length by default.

for loops seem to force the use of more checks, so I used while loops - which I'm less comfortable with.

Here's the code:

/**
 * TODO: If func counts out, let it return 0
 *  * Assume !isPalindrome (invert logic)
 */
function isPalindrome(S){
    var s = S
      , len = s.length
      , mid = len/2;
      , i = 0, j = len-1;

    while(i<mid){
        var l = s.charAt(i);
        while(j>=mid){
            var r = s.charAt(j);
            if(l === r){
                console.log('@while *', i, l, '...', j, r);
                --j;
                break;
            }
            console.log('@while !', i, l, '...', j, r);
            return 0;
        }
        ++i;
    }
    return 1;
}

var nooe = solution('neveroddoreven');  // even char length
var kayak = solution('kayak');  // odd char length
var kayaks = solution('kayaks');

console.log('@isPalindrome', nooe, kayak, kayaks);

Notice that if the loops count out, it returns true. All the logic should be inverted so that it by default returns false. I also used one short cut method String.prototype.charAt(n), but I felt OK with this as every language natively supports this method.

Answer 10

This function will remove all non-alphanumeric characters (punctuation, spaces, and symbols) and turn everything lower case in order to check for palindromes.

function palindrome(str){

    var re = /[^A-Za-z0-9]/g;
    str = str.toLowerCase().replace(re, '');
    return str == str.split('').reverse().join('') ? true : false;

}

Answer 11

Try this

isPalindrome = (string) => {
    if (string === string.split('').reverse().join('')) {
        console.log('is palindrome');
    }
    else {
        console.log('is not palindrome');
    }
}

isPalindrome(string)

Answer 12

String.prototype.isPalindrome = function isPalindrome() {

    const cleanString = this.toLowerCase().replace(/\s+/g, '');
    const cleanStringRevers = cleanString.split("").reverse().join("");

    return cleanString === cleanStringRevers;
}

let nonPalindrome = 'not a palindrome';
let palindrome = 'sugus';

console.log(nonPalindrome.isPalindrome())

console.log(palindrome.isPalindrome())

Answer 13

Here's a one-liner without using String.reverse,

const isPal = str => [...new Array(strLen = str.length)]
  .reduce((acc, s, i) => acc + str[strLen - (i + 1)], '') === str;

Answer 14

function palindrome(str) {
    var lenMinusOne = str.length - 1;
    var halfLen = Math.floor(str.length / 2);

    for (var i = 0; i < halfLen; ++i) {
        if (str[i] != str[lenMinusOne - i]) {
            return false;
        }
    }
    return true;
}

Optimized for half string parsing and for constant value variables.

Answer 15

I think following function with time complexity of o(log n) will be better.

    function palindrom(s){
    s = s.toString();
    var f = true; l = s.length/2, len = s.length -1;
    for(var i=0; i < l; i++){
            if(s[i] != s[len - i]){
                    f = false; 
                    break;
            }
    }
    return f;

    }

console.log(palindrom(12321));

Answer 16

Here's another way of doing it:

function isPalin(str) {
  str = str.replace(/\W/g,'').toLowerCase();
  return(str==str.split('').reverse().join(''));
}

Answer 17

Below code tells how to get a string from textBox and tell you whether it is a palindrome are not & displays your answer in another textbox

<html>
<head>
<meta charset="UTF-8"/>
<link rel="stylesheet" href=""/>

</head>  
<body>   
<h1>1234</h1>
<div id="demo">Example</div>
<a  accessKey="x" href="http://www.google.com" id="com" >GooGle</a>
<h1 id="tar">"This is a Example Text..."</h1>
Number1 : <input type="text" name="txtname" id="numb"/>
Number2 : <input type="text" name="txtname2" id="numb2"/>
Number2 : <input type="text" name="txtname3" id="numb3" />
<button type="submit"  id="sum" onclick="myfun()" >count</button>
<button type="button"  id="so2" onclick="div()" >counnt</button><br/><br/>
<ol>
<li>water</li>
<li>Mazaa</li>
</ol><br/><br/>
<button onclick="myfun()">TryMe</button>
    <script>
    function myfun(){
    var pass = document.getElementById("numb").value;
    var rev = pass.split("").reverse().join("");
    var text = document.getElementById("numb3");
    text.value = rev;
    if(pass === rev){
    alert(pass + " is a Palindrome");
    }else{
    alert(pass + " is Not a Palindrome")
    }
    }
    </script>
</body>  
</html>  

Answer 18

25x faster + recursive + non-branching + terse

function isPalindrome(s,i) {
 return (i=i||0)<0||i>=s.length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
}

See my complete explanation here.

Answer 19

The code is concise quick fast and understandable.

TL;DR

Explanation :

Here isPalindrome function accepts a str parameter which is typeof string.

1. If the length of the str param is less than or equal to one it simply returns "false".

2. If the above case is false then it moves on to the second if statement and checks that if the character at 0 position of the string is same as character at the last place. It does an inequality test between the both.

   str.charAt(0)  // gives us the value of character in string at position 0
   str.slice(-1)  // gives us the value of last character in the string.
   

If the inequality result is true then it goes ahead and returns false.

3. If result from the previous statement is false then it recursively calls the isPalindrome(str) function over and over again until the final result.

	function isPalindrome(str){
	
	if (str.length <= 1) return true;
	if (str.charAt(0) != str.slice(-1)) return false;
	return isPalindrome(str.substring(1,str.length-1));
	};


document.getElementById('submit').addEventListener('click',function(){
	var str = prompt('whats the string?');
	alert(isPalindrome(str))
});

document.getElementById('ispdrm').onsubmit = function(){alert(isPalindrome(document.getElementById('inputTxt').value));
}

<!DOCTYPE html>
<html>
<body>
<form id='ispdrm'><input type="text" id="inputTxt"></form>

<button id="submit">Click me</button>
</body>
</html>

Answer 20

    function palindrome(str) {
        var re = /[^A-Za-z0-9]/g;
        str = str.toLowerCase().replace(re, '');
        var len = str.length;
        for (var i = 0; i < len/2; i++) {
            if (str[i] !== str[len - 1 - i]) {
                return false;
            }
        }
        return true;
    }

Answer 21

Or you could do it like this.

var palindrome = word => word == word.split('').reverse().join('')

Answer 22

How about this one?

function pall (word) {

    var lowerCWord = word.toLowerCase();
    var rev = lowerCWord.split('').reverse().join('');

    return rev.startsWith(lowerCWord);
    }

pall('Madam');

Answer 23

str1 is the original string with deleted non-alphanumeric characters and spaces and str2 is the original string reversed.

function palindrome(str) {

  var str1 = str.toLowerCase().replace(/\s/g, '').replace(
    /[^a-zA-Z 0-9]/gi, "");

  var str2 = str.toLowerCase().replace(/\s/g, '').replace(
    /[^a-zA-Z 0-9]/gi, "").split("").reverse().join("");


  if (str1 === str2) {
    return true;
  }
  return false;
}

palindrome("almostomla");

Answer 24

Here is an optimal and robust solution for checking string palindrome using ES6 features.

const str="madam"
var result=[...str].reduceRight((c,v)=>((c+v)))==str?"Palindrome":"Not Palindrome";
console.log(result);

Answer 25

function isPalindrome(s) {
    return s == reverseString(s);
}

console.log((isPalindrome("abcba")));

function reverseString(str){
    let finalStr=""
    for(let i=str.length-1;i>=0;i--){
        finalStr += str[i]
    }
    return finalStr
}

Answer 26

Frist I valid this word with converting lowercase and removing whitespace and then compare with reverse word within parameter word.

function isPalindrome(input) {
    const toValid = input.trim("").toLowerCase();
    const reverseWord = toValid.split("").reverse().join("");
    return reverseWord == input.toLowerCase().trim() ? true : false;
}
isPalindrome("    madam ");
//true

Answer 27

This answer is easy to read and I tried to explain by using comment. Check the code below for How to write Palindrome in JavaScript. Step 1: Remove all non-alphanumeric characters (punctuation, spaces and symbols) from Argument string 'str' using replace() and then convert in to lowercase using toLowerCase(). Step 2: Now make string reverse. first split the string into the array using split() then reverse the array using reverse() then make the string by joining array elements using join() . Step 3: Find the first character of nonAlphaNumeric string using charAt(0). Step 4: Find the Last character of nonAlphaNumeric string using charAt(length of nonAlphaNumeric string - 1). Step 5: Use If condition to chack nonAlphaNumeric string and reverse string is same or not. Step 6: Use another If condition to chack first character of nonAlphaNumeric string is same to Last character of nonAlphaNumeric string.

function palindrome(str) {
  var nonAlphaNumericStr = str.replace(/[^0-9A-Za-z]/g, "").toLowerCase(); // output - e1y1e
 
  var reverseStr = nonAlphaNumericStr.split("").reverse().join(""); // output - e1y1e
  
  var firstChar = nonAlphaNumericStr.charAt(0); // output - e
  
  var lastChar = nonAlphaNumericStr.charAt(nonAlphaNumericStr.length - 1); // output - e
  
  if(nonAlphaNumericStr === reverseStr) {
    
    if(firstChar === lastChar) {
      return `String is Palindrome`; 
    }
  }
  return `String is not Palindrome`; 
}
console.log(palindrome("_eye"));

Answer 28

function check(txt)

{
  for (var i = txt.length; i >= 0; i--)

    if (txt[i] !== txt[txt.length - 1 - i])

      return console.log('not palidrome');
  return console.log(' palidrome');
}

check('madam');

Answer 29

Note: This is case sensitive

function palindrome(word)
{
    for(var i=0;i<word.length/2;i++)
        if(word.charAt(i)!=word.charAt(word.length-(i+1)))
            return word+" is Not a Palindrome";
    return word+" is Palindrome";
}

Here is the fiddle: http://jsfiddle.net/eJx4v/

Answer 30

I am not sure how this JSPerf check the code performance. I just tried to reverse the string & check the values. Please comment about the Pros & Cons of this method.

function palindrome(str) {
    var re = str.split(''),
        reArr = re.slice(0).reverse();

    for (a = 0; a < re.length; a++) {
        if (re[a] == reArr[a]) {
            return false;
        } else {
            return true;
        }
    }
}

JS Perf test