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Code Golf Challenge: Palindrome Detector

David on July 25, 2019

Challenge: Using Javascript, implement a method that detects if a string is a palindrome in as few characters as possible!

The goal of this exercise is to write as little code as possible. Readability and maintainability are not a concern!

Solutions in comments 👇👇

quickhand • Jul 26 '19

Short:

let p=s=>![...s].some((c,i,a)=>c!=a[a.length-i-1])

Shorter:

let p=s=>[...s].reverse().join("")==s

In both:

p("potato") //false
p("racecar") //true

(First works on palindromic arrays of characters as well as strings, second only on strings)

David • Jul 26 '19

Nice. Really like the approach with “some” in your first example

Massimo Artizzu • Jul 26 '19 • Edited

[...word].reverse().join('')==word

On the plus side, it works on emojis (.split('') doesn't), but not on complex ones (like 👨‍👩‍👧‍👦).

Siddharth • Nov 27 '21

you can shave off 2 chars by doing .join``

Shijie Zhou • Nov 15 '19

function palindrome(str) {
    var splitString = str.split("");
    var halfSplitString = splitString.length / 2;
    for (var i = 0; i < halfSplitString; i++) {
        if (str[i] !== str[str.length - i - 1]) {
            return false;
        }
        else {
            return true;
        }
    }
}

Aspen James • Jul 25 '19

Does this need to work on palindromes containing capitalization, spaces, punctuation, etc?

For example: "A man, a plan, a canal: Panama." - some would consider this a palindrome, some would

David • Jul 25 '19

Good question. Lets keep it simple: only checking for words. No handling for spaces, punctuation or numbers.

David • Jul 26 '19 • Edited

Nice! You can make the function even smaller if you just returned



Array.from(word)
.reverse()
.join("") === word

Hao • Jul 26 '19

Even shorter

word.split('').reverse().join('')==word

𝐍𝐚𝐭𝐚𝐥𝐢𝐞 𝐝𝐞 𝐖𝐞𝐞𝐫𝐝 • Jul 26 '19

This wouldn't work for "Arara" would it? It'd need some sort of case normalisation.

Also, what about "Taco cat".

quickhand • Jul 26 '19 • Edited

The issuer of the challenge, David, said:
"Lets keep it simple: only checking for words. No handling for spaces, punctuation or numbers."

But if we're handling capitals and spaces, how about:

let p=s=>(t=>[...t].reverse().join("")==t)(s.toLowerCase().replace(/\s/g,""))

Here,

p("Taco cat") //returns true,
p("Arara") //returns true

Another slight mod and we can ignore all non-letter chars (numbers, punctuation, etc.):

let p=s=>(t=>[...t].reverse().join("")==t)(s.toLowerCase().replace(/[^A-z]/g,""))

p("A man, a plan, a canal: Panama.") // returns true

David • Jul 25 '19 • Edited

let p = (w) => { let i = 0,c = w[i],l=w.length - 1;while(c){if(c !== w[l-i]){return false}c=w[++c]}return true}

Lorraine Lee • Jul 28 '19

function p(a) {while (a.length>1&&(v=a.shift()==a.pop())); return v;}

Siddharth • Nov 27 '21

This is the shortest IMO:

s=>[...s].reverse``.join``==s

With normalization:

s=>{s=s.replace(/\s/g,'').toLowerCase();return[...s].reverse``.join``==s}