I try to carve out about 2 hours a day to work on code problems. In addition to solving them, I find it really useful to write a brief expanation of my process. It gets me in the habit of putting words to my coding decisions, a skill that interviewers want to see when you’re working through a technical problem. Writing out the approach also helps with remembering certain strategies when encountering similar problems down the road.
Compare the Triplets
Lula and Jack each wrote their own sitcom pilot. They asked a friend to rate the two pilots, awarding points on a scale from 1 to 100 for three categories: originality, humor, and storyline.
The rating for Lula’s sitcom, “Dinosaur Wall Street” is the triplet a = (a[0], a[1], a[2]), and the rating for Jack’s sitcom, “Small Town Napkin Factory” is the triplet b = (b[0], b[1], b[2]).
Your 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].
- If a[i] > b[i], then Lula is awarded 1 point.
- If a[i] < b[i], then Jack is awarded 1 point.
- If a[i] = b[i], then neither person receives a point.
Comparison points are the total points a person earned.
Given a and b, determine their respective comparison points.
Example
a = [1, 2, 3] b = [3, 2, 1]
- For elements 0, Jack 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 Lula receives a point.
The return array is [1, 1] with Lula’s score first and Jack’s second.
Function Description
Complete the function compareTriplets in the editor below.
compareTriplets has the following parameter(s):
- int a[3]: Lula’s challenge rating
- int b[3]: Jack’s challenge rating
Return
- int[2]: Lula’s score is in the first position, and Jack’s score is in the second.
Input Format
The first line contains 3 space-separated integers, a[0], a[1], and a[2], the respective values in triplet a. ** The second line contains 3 space-separated integers, b[0], b[1], and b[2], the respective values in triplet b.
Constraints
- 1 ≤ a[i] ≤ 100
- 1 ≤ b[i] ≤ 100
Sample Input 0
[5, 6, 7]
[3, 6, 10]
Sample Output 0
[1, 1]
Explanation 0
In this example:
- a = (a[0], a[1], a[2]) = (5,6,7)
- b = (b[0], b[1], b[2]) = (3, 6, 10)
Now, let’s compare each individual score:
- a[0] > b[0], so Lula receives 1 point.
- a[1] = b[1], so nobody receives a point.
- a[2] < b[2], so Jack receives 1 point.
Lula’s comparison score is 1, and Jack’s comparison score is 1.
Thus, we return the array [1, 1].
Sample Input 1
[17, 28, 30]
[99, 16, 8]
Sample Output 1
[2, 1]
Explanation 1
Comparing the 0th elements, 17 < 99 so Jack receives a point.
Comparing the 1st and 2nd elements, 28 > 16 and 30 > 8 so Lula receives two points.
The return array is [2, 1].
Solution and Explanation
*There is rarely one way to solve a problem and this is simply one solution. There are probably many solutions better than this and a few that are worse. *
The task is to write a function, compareTriplets() that accepts two arrays as an argument (Lula and Jack’s individual sitcom-rating triplets) and returns an array of their respective comparision points. Long story short, compare the two arrays, awarding a point to the person whose rating is higher for each category. Then return the “score.”
function compareTriplets(a, b) {
// Compare the triplets
// Return the score
}
I know I’ll need to eventually return an array of the comparison points, so my first step is to declare a variable score and assign it an inital value.
function compareTriplets(a, b) {
let score = [0,0]
}
I also know what the function needs to return.
function compareTriplets(a, b) {
let score = [0,0]
// compare the two arrays
return score
}
The comparison itself is fairly straightforward - higher rating in each category (each index) gets a point, equal ratings do nothing. Iterating through the arrays using a for loop, I compare each index. If the two integers are not equal, I award a point to the correct score index. Returning the score array after the loop completes.
function compareTriplets(a, b) {
let score = [0,0]
for (let i=0; i < a.length; i++) {
if (a[i] !== b[i]){
a[i] > b[i] ? score[0] += 1 : score[1] += 1
}
}
return score
}
At this point I have working solution. I can move onto something else or take the time to re-examine my solution, looking for alternate approaches or potential improvements. This is putting in the work.