Discussion on: Keeping Your Code Simple

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Sometimes, there's an even simpler solution:

const strs = ['hello', 'goodbye', 'farewell', 'aloha'];

strs.sort((str1, str2) => str2.length - str2.length);

const longestString = strs[0];

Nando • Aug 17 '18

I don't think this is a simpler solution, albeit from a logical point of view: it may be understood only considering the outcome of the sorting process, then getting the first item as the one with the minor size...

Lars Gyrup Brink Nielsen • Aug 17 '18 • Edited

If you turn that into a reusable function, you will be mutating the array parameter, i.e. sorting it in-place. You would have to clone the array before sorting it to prevent side effects from passing an array into the function.

Also, you made a mistake. You compare str2.length to str2.length (both are str2).

Use this instead

function longest(xs) {
  // Shallow clone
  xs = [...xs];
  xs.sort((a, b) => b.length - a.length);

  return xs[0];
}

Or with reusable functions

function compareLength(a, b) {
  return b.length - a.length;
}

function head([first, ...rest]) {
  return first;
}

function longest(xs) {
  xs = shallowClone(xs);
  xs.sort(compareLength);

  return head(xs);
}

function shallowClone(xs) {
  return [...xs];
}

Alex Lohr • Aug 17 '18

You're right.

theMightiestT • Aug 16 '18

i thought the same... and if you have strings with numbers, then add a return (a-b) or something...

Evaldas Buinauskas • Aug 16 '18

Seems that sort has worse performance compared to reduce. Reduce is O(n) while sort is O(n long)

Alex Lohr • Aug 17 '18

Yes, the performance is worse. But unless you need the solution to scale to arrays with more than a few thousand words, this will hardly be an issue.

Evaldas Buinauskas • Aug 17 '18

True. Just wrote that because it wasn't mentioned anywhere and someone might care about it.

theMightiestT • Aug 17 '18

valid point but I agree with Alex... not sure it would really be a thing in an implementation like this