'Break'-ing the Palindrome Checker Algorithm

In a previous article, I addressed the palindrome algorithm using recursion.

A Few Exercises for Learning Recursion

Clark Johnson ・ Jun 8 '20 ・ 3 min read

#javascript

A few days ago, I came across a solution using an iterator to get the job done.

function palindrome2(str) {
    return str
        .split('')
        .every(
            (char, i) => char === str[str.length - 1 - i]
        )
}

This idea utilizes a built-in Javascript iterator 'every', which returns true when a condition is met by every item in the array.

GIF

Take the word 'ROTATOR'. It's definitely a palindrome, and we can tell by comparing the first letter to the last letter, then the second letter to the next-to-last letter, and continue until every letter has been tested. If every letter has a match on the opposite side of the word, then we have a palindrome.

The .every function will get the job done. However, we're doing a little too much work. We should be able to stop testing for matching equalities once we reach the halfway point. After that, every check is redundant. Unfortunately, using .every doesn't provide a way to break out of iteration and continue.

Thanks to ES6, we have another solution. Using the for... of construct to iterate gives the option to take advantage of the break keyword.

function palindrome(str) {
    let valid = true;
    const arr = str.split('')
    for (const [i, char] of arr.entries()) {
        if (i > (arr.length / 2))
            break;
        else if (char !== arr[arr.length - i - 1]) {
            valid = false;
            break
        }
    }

    return valid
}

This solution isn't quite as elegant as the first, but it is definitely more efficient. Execution will now cease if either the palindrome test fails on a character or we reach the halfway point. The addition control removes redundancy and reduces the execution time.

More information about for...of is available on MDN.

Happy coding!

Cover Photo by Justin Main on Unsplash