DEV Community

Reverse a String - Four JavaScript Solutions

Sarah Chima on July 17, 2019

This article was originally published on my blog The reverse a string problem is a common algorithm problem. In this article, we will consider fou...

Read full post

Bruce Axtens • Jul 18 '19 • Edited

Actually, you could take the recursion along a bit further and make it a prototype of String, viz:

String.prototype.flip = function () { return (this.length === 1) ? this : this.substr(1).flip() + this.substr(0, 1); }; var original = "lewd i did live - evil did i dwel"; var reversed = original.flip();

Someone'll beat me to a blog post, likely.

Abraham Williams • Jul 18 '19

It's generally recommended that you should not extend native prototypes.

developers.google.com/web/updates/...

Mark • Jul 18 '19

shakes fist at MooTools

hidden_dude • Jul 17 '19 • Edited

Wouldn't solution #2 (and maybe some of the others) cause an O(n² ) problem?

for (let character of str) { reverseString = character + reverseString; }

If str has N elements, and you do character + reverseString each time you get n*(n+1)/2 characters copied (ie. O(n² )).

Whereas option#1 presumably uses only O(n) functions so it should be much faster as data grows and less stress on the GC.

I think it would be best to compare these with benchmarks (with large examples). Because I've found that the one place where code can get super slow is with successive concatenation in a for loop (as in option#2).

Sarah Chima • Jul 18 '19

Thanks for the suggestion and the analysis. In subsequent posts, I try to compare the solutions based on their performance.

Theofanis Despoudis • Jul 18 '19

What about an old classic way:

function reverse(str) { let result = str.split(''); for (let i = 0, j = str.length - 1;i < j; i+=1, j-=1) { result[i] = str[j]; result[j] = str[i]; } return result.join(''); }

Sarah Chima • Jul 18 '19

This is a good improvement to the solution 2. It reduces the loop time by half. Thanks for sharing.

Nathanael Demacon • Jul 18 '19

.split('') can be shortened:

[...string].reverse().join('')

Rhys Brett-Bowen • Jul 18 '19

Also good because split will fail badly for some utf8 things like emojis

Bruce Axtens • Jul 18 '19

So after doing .call(), let's do .apply(), viz

Array.prototype.flip = function () { for (var i = 0; i < this.length / 2; i++) { var tmp = this[this.length - i - 1]; this[this.length - i - 1] = this[i]; this[i] = tmp; } return this; } var original = "lewd i did live - evil did i dwel"; var reversed = [].flip.apply(original.split("")).join("");

Okay, maybe it's time to do some real work now.

Bruce Axtens • Jul 19 '19 • Edited

This recursion thing is pretty cool. Here's another approach both as a String.prototype and as a standalone function:

String.prototype.flip2 = function() { if (1 >= this.length) return this; return ( this.substr(-1) + this.substr(1, this.length - 2).flip2() + this.substr(0, 1) ); }; function flip2(string) { if (1 >= string.length) return string; return ( string.substr(-1) + flip2(string.substr(1, string.length - 2)) + string.substr(0, 1) ); } var x = "lewd i did live - evil did i dwel".flip2(); var y = flip2("lewd i did live - evil did i dwel");

The approach is to take the first character and put it last, take the last character and put it first, and do that to the stuff in the middle.

It's been through the Closure Compiler, thus the 1 >= this.length.

Bruce Axtens • Jul 18 '19 • Edited

I'm not sure what you'd call this solution, but it does reverse the string:

var times = function (n, fn /*, args*/) { var args = [].slice.call(arguments).slice(1); for (var i = 0; i < n; i++) { fn.call(this, i, args); } return args[2]; } var original = "lewd i did live - evil did i dwel"; var result = ""; var reversed = times(original.length, function (i, args) { args[2] = args[1].substr(i, 1) + args[2]; }, original, result);

That's the first time I've written code that uses .call(). Won't be the last time!

Bruce Axtens • Jul 19 '19

Going nuts here. More ideas, and for something as mundane as a reverse function. Monomania?

function flip3(string) { var result = Array(string.length); for (var i = string.length - 1, j = 0; i >= 0; i--, j++) { result[j] = string.charAt(i); } return result.join(""); } function flip4(string) { var result = Array(string.length).fill(1); result.map(function (item,index) { var rhs = string.length - 1 - index; result[index] = string.charAt(index); }); return result.join(""); }

Your mileage may vary on flip4's .fill() call. I'm having to use polyfills (long story) and if I don't fill the Array, .map() assumes that there's nothing there to work on.

I do have a couple of ideas yet to exorcise.

quickhand • Jul 19 '19

Short, sweet:

[...str].map((_,i,arr) => arr[arr.length-1-i])

Bruce Axtens • Jul 19 '19

I'd forgotten about that third parameter into map. Short, sweet, superb!

Wernich ️ • Jul 19 '19

first thing i thought of was reduceRight (works in the opposite direction to reduce), with no split or join :)

Array.from(str).reduceRight((acc, ch) => acc.concat(ch));

Martín Granados García • Jul 17 '19

Nice idea for blog posts! It is helpful for people wanting to learn and helpful for you to practice the algos! 💯

Sarah Chima • Jul 18 '19

Thank you, Martin.

Felix Wittmann • Jul 23 '19 • Edited

certainly not very fast, but I'd like destructuring assignment and recursion :)

let reverse = ([h,...r]) => r.length>0 && reverse(r)+h || h;

Sriram Narasimhan • Jul 19 '19

Another solution using for loop, but only iterates half the time.

function reverse(str){ let reversedStr = str const length = reversedStr.length for(i=0; i< length/2; i++){ const endIndex = length - (i+1) if(endIndex > i){ const pre = reversedStr.substring(0, i) const beginChar = reversedStr.substring(i, i+1) const unsorted = reversedStr.substring(i+1,endIndex) const endChar = reversedStr.charAt(endIndex) const post = reversedStr.substring(endIndex+1) //An overkill of swap ! reversedStr = pre+endChar+unsorted+beginChar+post } } return reversedStr } console.log(reverse("1234567")) console.log(reverse("123456")) console.log(reverse("abracadabra")) console.log(reverse("esrever"))