Compare Array [Triplets] - ios

Question

I'm trying to complete this
a = [5, 6, 7]
b = [3, 6, 10]
Now, let's compare each individual score:
a[0] > b[0], so Alice receives point.
a[1] == b[1], so nobody receives a point.
a[2] < b[2] , so Bob receives point.
Alice's comparison score is 1, and Bob's comparison score is 1. Thus, we have to print 1 1 (Alice's comparison score followed by Bob's comparison score) on a single line.
I did it by this way:
var a = [5, 6, 7]
var b = [3, 6, 10]
let pointsAlice = (a[0] > b[0] ? 1: 0) + (a[1] > b[1] ? 1: 0) + (a[2] > b[2] ? 1: 0)
let pointsBob = (a[0] < b[0] ? 1: 0) + (a[1] < b[1] ? 1: 0) + (a[2] < b[2] ? 1: 0)
print("\(pointsAlice) \(pointsBob)")
But it is showing as an incorrect answer(Hackerrank).
Is my way correct or is there any other way to solve this ?

Answer 1

Please try this ...
let a = readLine()!.components(separatedBy: " ").flatMap { Int($0) }
let b = readLine()!.components(separatedBy: " ").flatMap { Int($0) }
let aResult = zip(a, b).filter(>).count
let bResult = zip(b, a).filter(>).count
//Final Result Status
print(aResult, bResult)

Answer 2

I guess u concentrated on explanation more :p
Try this:
let arrA = readLine()!.components(separatedBy: " ").map{ Int($0)! }
let arrB = readLine()!.components(separatedBy: " ").map{ Int($0)! }
var aliceScore = 0
var bobScore = 0
for index in 0...2{
if (arrA[index] > arrB[index]){
aliceScore += 1
}
if (arrA[index] < arrB[index]){
bobScore += 1
}
}
print("\(aliceScore)" + " " + "\(bobScore)")

Answer 3

This was my solution. Trying to learn how the reduce func works.
func compareTriplets(a: [Int], b: [Int]) -> [Int] {
var total = [0,0]
for x in 0..<a.count {
if a[x] > b[x] {
total[0] += 1
} else if a[x] < b[x] {
total[1] += 1
}
}
return aTotal
}

Answer 4

Why not just use higher order functions like reduce?
let a = [5, 6, 7]
let b = [3, 6, 10]
let pointsAlice = zip(a, b).reduce(0, combine: { $0 + ($1.0 > $1.1 ? 1 : 0) } )
let pointsBob = zip(b, a).reduce(0, combine: { $0 + ($1.0 > $1.1 ? 1 : 0) } )
print("\(pointsAlice) \(pointsBob)") // 1 1

Answer 5

Do not hard code like this
(a[0] > b[0] ? 1: 0) + (a[1] > b[1] ? 1: 0) + (a[2] > b[2] ? 1: 0)
Imagine if the array has 1,000 or more elements!!!
I think the simplest solution is:
var alicePoints = 0
var bobPoints = 0
for i in 0..<A.count {
if A[i] > B[i] {
alicePoints += 1
} else if A[i] < B[i] {
bobPoints += 1
}
}
print("\(alicePoints) \(bobPoints)")
And the more elegant solution is the one from #koropok above (I upated for Swift 3.x)
let bobPoints = zip(B, A).reduce(0) { $0 + ($1.0 > $1.1 ? 1 : 0) }
let alicePoints = zip(A, B).reduce(0) { $0 + ($1.0 > $1.1 ? 1 : 0) }
print("\(alicePoints) \(bobPoints)")

Answer 6

Here is the updated Answer for comparing Triplets
Let's say if we have
var a = [5, 6, 7]
var b = [3, 6, 10]
Then
// compareTriplets function
func compareTriplets(a: [Int], b: [Int]) -> [Int] {
let bobPoints = zip(b, a).reduce(0) { $0 + ($1.0 > $1.1 ? 1 : 0) }
let alicePoints = zip(a, b).reduce(0) { $0 + ($1.0 > $1.1 ? 1 : 0) }
print("Points earned by Alice and Bob : \(alicePoints) \(bobPoints)")
return [alicePoints, bobPoints]
}

Answer 7

Try this:
return [zip(a, b).reduce(0) {$0 + ($1.0 > $1.1 ? 1 : 0)},
zip(b, a).reduce(0) {$0 + ($1.0 > $1.1 ? 1 : 0)}]

Answer 8

In python, without considering the constraints proposed(ab >= 1 & ab<=100):
def compareTriplets(a, b):
i = 0
score = [0,0]
while i < 3:
if a[i] > b[i]:
score[0] = score[0] + 1
elif a[i] < b[i]:
score[1] = score[1] + 1
i = i + 1
return score

Answer 9

import java.util.*.;
public class Solution {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int klice=0;
int bob=0;
int a[]=new int[n];
int b[]=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
for(int i=0;i<n;i++){
b[i]=sc.nextInt();
}
for(int i=0;i<n;i++){
if(a[i]>b[i]){
klice++;
} else if(b[i]>a[i]){
bob++;
}
}
System.out.println(klice +" "+ bob);
}
}