re: Javascript Sock Merchant Challenge - Solution 2 VIEW POST

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re: An alternative solution: const a = [10, 20, 20, 10, 10, 30, 50, 10, 20] // [[10, 10, 10, 10], [20, 20, 20], [30], [50]] // => 3 complete pai...
 

Hello John, I have definitely enjoyed reading your post and have learned a few nice tricks from it. My issue though with the solution will be readability first. It really takes a minute to grasp what is going on here. The second one is that a lot of transforms (map chain) are performed here where they probably could be taken care of by one computation reducing the complexity and improving the performance. Running against a simple benchmark this solution is over 300% slower than the proposed one.
I will host a webinar probably next week that will focus on why this second approach might be more suitable for this type of challenges. It will mean the world to me if you could attend and add to the discussion

 

Hello Ady, thanks for the reply.

I recognize that my solution must look a bit strange on first sight. I was reading on MDN recently about the draft proposal for the pipeline operator aka |> to be added to JavaScript(as yet unsupported).

I also had passed sometime on the egghead.io site
re: egghead.io/courses/professor-frisb... and was intrigued how .map is used here to produce a topdown workflow composing(or piping) one function's result to the next function. What a shame that, as you have discovered, JS is not performant in this way, because, I for one, think this is an elegant and clean workflow that I hope, one day, will be worth using.

Of course speed of execution is always an important issue and so for the moment, atleast, we fall back to the roots of JavaScript and use it's for, for of, while, etc loops. Shame, because I really do like a more declarative style of coding and here I was merely experimenting.

Perhaps you will forgive a quick hack to make my solution a tad more readable(see code below) by placing a pipe method on the Array prototype. Sorry, I can't do anything about JS performance, but I think somewhat less than 10,000 socks to sort would be favourable; that sock merchant needs to keep his house in closer order I think!

Incidentally, please excuse me if I decline the webinar participation; it really is not my thing and you may take this reply as my input to that.

Have a great day.


if(!Array.prototype.pipe) Array.prototype.pipe = Array.prototype.map
const reduce = (fn, init) => xs => Array.prototype.reduce.call(xs, fn, init)

const a = [10, 20, 20, 10, 10, 30, 50, 10, 20]
// [[10, 10, 10, 10], [20, 20, 20], [30], [50]]
// => 3 complete pairs (& 3 odd socks)

const group = (p,c) =>
  (c in p)
    ? {...p, [c]: p[c] + 1}
    : {...p, [c]: 1}

// Note: ~~ is Math.floor in js.
const sumPairs = (acc, x) => acc + ~~(x/2)

// Note: .pipe is function composition. Pipes the result of a function to the next.
// Visit https://egghead.io/lessons/javascript-linear-data-flow-with-container-style-types-box
const completePairs = (n, ary) =>
  [ary]
    .pipe(reduce(group, {}))
    .pipe(Object.values)
    .pipe(reduce(sumPairs,0))
    .pop()

console.log(completePairs(9, a))
// => 3

As a final request, I would be interested in what benchmark timings (in milliseconds) you achieved for my soloution vs your solution for a test array of 10,000 integers.

 

A somewhat faster solution.


const count = (ofValue, ary) => {
  let i = ary.length
  let c = 0
  while(i--){ if(ary[i] === ofValue) {c++} }
  return c
}

const completePairs = (n, Ary) =>
  Array
    .from(new Set(Ary))
    .reduce((acc, v) => acc + ~~(count(v, Ary) / 2), 0)

My hope is that this is more readable and fast enough to satisfy most users.

I will take a closer look and since you don’t like to be put on the spot πŸ˜€ I think I want to have our discussion transcend the comments section with an article that captures the exchange.
In the meantime I’ll run a benchmark on it but as far as readability this seems easier to digest IMHO.
I have started to write in a more functional way and isolating the arity for flexible composition. I think in the end it comes down to choosing the right tool and style for the right job.

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