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Who doesnt have a pair answer #15

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StudentDawid
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#5 Answer

bear-in-a-box
bear-in-a-box suggested changes Oct 22, 2019
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This algorithm needs an explanation on why it's working. JSDocs with typings for better writing experience would be great to see as well.

Algorithms/pair.js
const tmp = [1, 2, 3, 4, 5, 4, 3, 2, 1];
const tmpAnswer = 5

const aloneAtParty = (tmp) => {
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Missing JSDocs on parameter and return typing

Algorithms/pair.js
const tmpAnswer = 5

const aloneAtParty = (tmp) => {
const a = tmp.reduce((acc, e) => {return acc+e}, 0);
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There's no need to do {return X} here. Instead, (acc, e) => acc + e would be sufficient. Also, we don't know what "a" variable name means.

Algorithms/pair.js

const aloneAtParty = (tmp) => {
const a = tmp.reduce((acc, e) => {return acc+e}, 0);
const b = [...new Set(tmp)].reduce((acc, e) => {return acc+e}, 0);
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Same thing regarding variable naming and code styling as above.

Algorithms/pair.js
const aloneAtParty = (tmp) => {
const a = tmp.reduce((acc, e) => {return acc+e}, 0);
const b = [...new Set(tmp)].reduce((acc, e) => {return acc+e}, 0);
return b-(a-b);
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This needs a little bit of an explanation comment.

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@StudentDawid @bear-in-a-box