I was practicing in Hackerrank JavaScript problems. I found one test which is called `Compare the triplets`

. This is the problem:

a = [1, 2, 3] b = [3, 2, 1] For elements *0*, Bob is awarded a point because a[0] . For the equal elements a[1] and b[1], no points are earned. Finally, for elements 2, a[2] > b[2] so Alice receives a point. The return array is [1, 1] with Alice's score first and Bob's second.

**I found the solution like this:**

let a = [17, 28, 30]; let b = [99, 16, 8]; function compareTriplets(a, b) { let scoreboard = [0, 0]; for (let i = 0; i < a.length; i++) { if (a[i] > b[i]) scoreboard[0]++ else if (a[i] < b[i]) scoreboard[1]++ } return scoreboard } compareTriplets(a, b)

I wanted to convert the ForLoop into `ForEach`

method. But I could not find the way to do that.

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## Answer

This ain’t what you asked for, but let me show you something:

function compareTriplets(a, b) { return [ (a[0] > b[0]) + (a[1] > b[1]) + (a[2] > b[2]), (a[0] < b[0]) + (a[1] < b[1]) + (a[2] < b[2]) ] }

or, less noise:

function compareTriplets([a, b, c], [d, e, f]) { return [ (a > d) + (b > e) + (c > f), (a < d) + (b < e) + (c < f) ] }

simpler, faster and also shorter.

I mean, it’s literally called “compare triplets”. There ain’t any dynamic length or something; and the loop is short. You can easily unroll the loop.

let a = [17, 28, 30]; let b = [99, 16, 8]; function compareTriplets([a, b, c], [d, e, f]) { return [ (a > d) + (b > e) + (c > f), (a < d) + (b < e) + (c < f) ] } console.log(compareTriplets(a, b));