My approach to solving HackerRank’s Compare the Triplets code challenge.
Alice and Bob each created one problem for HackerRank. A reviewer rates the two challenges, awarding points on a scale from 1 to 100 for three categories: problem clarity, originality, and difficulty.
The rating for Alice's challenge is the triplet a = (a[0], a[1], a[2]), and the rating for Bob's challenge is the triplet b = (b[0], b[1], b[2]).
The task is to find their comparison points by comparing a[0] with b[0], a[1] with b[1], and a[2] with b[2].
Comparison points is the total points a person earned.
Given a and b, determine their respective comparison points.
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.
int[2]
: Alice's score is in the first position, and Bob's score is in the second.
15 6 723 6 10
11 1
We start by creating a few variables to hold the comparison results.
Both arrays are equal in length, so we can do a for loop based on any array size. Then during each loop, we compare values of both arrays, and if a[i]
is bigger than b[i]
, then we bump aScore
by 1
. Else we bump the score for bScore
.
Then we return the results of the scores as an array.
1function compareTriplets(a, b) {2 let aScore = 0,3 bScore = 0;4
5 for (let i = 0; i < a.length; i++) {6 if (a[i] > b[i]) {7 aScore++;8 } else if (a[i] < b[i]) {9 bScore++;10 }11 }12
13 return [aScore, bScore];14}
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