Compare the Triplets | HackerRank Solution in Java with Explanation

Java Solution for Compare the Triplets HackerRank Problem

Problem Description:

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].  

• If a[i] > b[i], then Alice is awarded 1 point. 
• If a[i] < b[i], then Bob is awarded 1 point.  
• If a[i] = b[i], then neither person receives a point.

Comparison points is the total points a person earned.

Given a and b, determine their respective comparison points. 

See Full description :

• Compare the Triplets Problem Description 

Sample Explanation :

a = [1, 2, 3]
b = [3, 2, 1]

1. For elements 0, Bob is awarded a point because a[0] < b[0] .
2. For the equal elements a[1] and b[1], no points are earned.
3. 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.

Lets see solution

Solution 1

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    public static List<Integer> compareTriplets(List<Integer> a, List<Integer> b) {
        
         // Define and initialize list
         List<Integer> points = new ArrayList();
         points.add(0);
         points.add(0);
        
        // Compare all points of alice "a" and bob "b", 
        // and store result into list    
        for (int i = 0; i < a.size(); i++) {
            if (a.get(i) > b.get(i)) {
                points.set(0, points.get(0) + 1);
            } else if (a.get(i) < b.get(i)){
                points.set(1, points.get(1) + 1);
            }
        }
        return points;
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(
            new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new 
            BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        List<Integer> a = Stream.of(bufferedReader.readLine()
            .replaceAll("\\s+$", "").split(" "))
            .map(Integer::parseInt)
            .collect(toList());

        List<Integer> b = Stream.of(bufferedReader.readLine()
            .replaceAll("\\s+$", "").split(" "))
            .map(Integer::parseInt)
            .collect(toList());

        List<Integer> result = Result.compareTriplets(a, b);

        bufferedWriter.write(result.stream()
                .map(Object::toString)
                .collect(joining(" "))
            + "\n"
        );

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Solution explanation:

• We have two list of Integers a and b. We have to return list of  Integers, inside that one value is contains alice's score and second is bob's score.
• Define one list initialize with ArrayList. Add two 0 integers value with .add() method of list Interface.
• Traverse through given list, from 0 to list size. we can use any list from given parameter because both list contains same size.
  
• Now we compare both alice and bob's score one by one. 
• If Alice's score is greater than Bob's score, we are increment Alice's score by one using .set() method of list interface.
• In Else if condition, we are checking If Alice's score is less than Bob's score (means Bob's score is greater than alice's score), we are increment Bob's score by one using .set() method of list interface.  
• Last, return list of score.

Output explanation :

a = [17, 28, 30]
b = [99 ,16 , 8]

• i = 0, points = [0, 0]
• If Condition | a.get(0) > b.get(0) | 17 > 99 becomes false.
• Else If Condition |  a.get(0) < b.get(0) | 17 < 99 becomes true.
• points.set(1, points.get(0) + 1) |  points = [0, 1].
  
  
• i = 1, points = [0, 1]
• If Condition | a.get(0) > b.get(0) | 28 > 16 becomes true.
• points.set(0, points.get(0) + 1) |  points = [1, 1].
  
  
• i = 2, points = [1, 1]
• If Condition | a.get(0) > b.get(0) | 30 > 8 becomes false.
• points.set(0, points.get(0) + 1) |  points = [2, 1].
  
  
• points = [2, 1] | alice's score is 2 and bob's score is 1.

Lets see another solution.

Solution 2

public static List<Integer> compareTriplets(List<Integer> a, List<Integer> b) {
    
    int alice_score, bob_score = 0;
    List<Integer> points = new ArrayList<Integer>();
                
    for (int i = 0; i < 3; i++) {
        if (a.get(i) > b.get(i)) {
            alice_score++;
        }
        else if (a.get(i) < b.get(i)) {
            bob_score++;
        }
    }
    
    points.add(alice_score);
    points.add(bob_score); 
    return points;
}

Solution explanation :

This solution is same as above with some change. 

• We take two int variable name, alice_score and bob_score.
• When alice's point is greater than bob's point, we are incrementing alice_score by 1. If bob's point is greater than alice, then we are incrementing bob_score by 1.
• Last, we are adding both score into list and return list. 

Happy Coding.

See other HackerRank solutions :

• Java Interface HackerRank Solution with Explanation 
• HackerRank Solution for Array Left Rotation