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# Compare the Triplets Hacker Rank Solution – Queslers

## Problem: Compare the Triplets Hacker Rank Solution

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 clarityoriginality, and difficulty.

The rating for Alice’s challenge is the triplet a = (a, a, a), and the rating for Bob’s challenge is the triplet b = (b, b, b).

The task is to find their comparison points by comparing a with ba with b, and a with b.

• 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.

Example

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

• For elements *0*, Bob is awarded a point because a .
• For the equal elements a and b, no points are earned.
• Finally, for elements 2a > b so Alice receives a point.

The return array is [1, 1] with Alice’s score first and Bob’s second.Function DescriptionComplete the function compareTriplets in the editor below.compareTriplets has the following parameter(s):int a: Alice’s challenge ratingint b: Bob’s challenge ratingReturn

• int: Alice’s score is in the first position, and Bob’s score is in the second.

Input Format

The first line contains 3 space-separated integers, aa, and a, the respective values in triplet a.
The second line contains 3 space-separated integers, bb, and b, 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 1Alice 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, a, a), and the rating for Bob's challenge is the triplet b = (b, b, b).

The task is to find their comparison points by comparing a with b, a with b, and a with b.

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.

Example

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

Function Description

Complete the function compareTriplets in the editor below.

compareTriplets has the following parameter(s):

int a: Alice's challenge rating
int b: Bob's challenge rating
Return

int: Alice's score is in the first position, and Bob's score is in the second.
Input Format

The first line contains 3 space-separated integers, a, a, and a, the respective values in triplet a.
The second line contains 3 space-separated integers, b, b, and b, 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,a,a)=(5,6,7)
b=(b,b,b)=(3,6,10)
Now, let's compare each individual score:

a>b, so nobody receives a point.
Alice's comparison score is 1 , and Bob's comparison score is 1  . Thus, we return the array [1,1].
```

### Compare the Triplets Hacker Rank Solution Using Python

``````#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the compareTriplets function below.
def compareTriplets(a, b):
nas = []
x=0
y=0
for i in range(0,len(a)):

if a[i]>b[i]:
x=x+1
elif b[i]>a[i]:
y=y+1
else:
continue
nas.append(x)
nas.append(y)
return nas

if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

a = list(map(int, input().rstrip().split()))

b = list(map(int, input().rstrip().split()))

result = compareTriplets(a, b)

fptr.write(' '.join(map(str, result)))
fptr.write('\n')

fptr.close()``````

### Compare the Triplets Hacker Rank Solution Using Java

``````import java.io.*;
import java.util.*;

public class Solution {

public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */

Scanner scan = new Scanner(System.in);
//    	int N = scan.nextInt();
int[] alice = new int;
int[] bob = new int;
int a=0, b=0;
for(int i=0;i<3;i++)
alice[i]=scan.nextInt();
for(int i=0;i<3;i++)
bob[i]=scan.nextInt();
for(int i=0;i<3;i++)
if(alice[i]>bob[i])
a++;
else if(alice[i]<bob[i])
b++;
System.out.println(a+" "+b);
scan.close();
}
}``````

### Compare the Triplets Hacker Rank Solution Using C++

``````#include <bits/stdc++.h>

template<typename T> T gcd(T a, T b) {
if(!b) return a;
return gcd(b, a % b);
}
template<typename T> T lcm(T a, T b) {
return a * b / gcd(a, b);
}

template<typename T> void chmin(T& a, T b) { a = (a > b) ? b : a; }
template<typename T> void chmax(T& a, T b) { a = (a < b) ? b : a; }
int in() { int x; scanf("%d", &x); return x; }

using namespace std;

#ifdef ONLINE_JUDGE
#define debug(args...)
#else
#define debug(args...) fprintf(stderr,args)
#endif

typedef long long Int;
typedef unsigned long long uInt;
typedef unsigned uint;

int A, B;

int main(void) {
int ia = 0;
int ib = 0;

for (int i = 0; i < 3; i++) {
cin >> A[i];
}
for (int i = 0; i < 3; i++) {
cin >> B[i];

if (A[i] < B[i]) {
ib += 1;
} else if (A[i] > B[i]) {
ia += 1;
}
}

cout << ia << " " << ib << "\n";
return 0;
}``````

### Compare the Triplets Hacker Rank Solution Using C

``````#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int a0;
int a1;
int a2;
scanf("%d %d %d",&a0,&a1,&a2);
int b0;
int b1;
int b2;

int a_score = 0;
int b_score = 0;
scanf("%d %d %d",&b0,&b1,&b2);
if (a0 > b0)
a_score++;
else if (a0 < b0)
b_score++;
else{}
//no op

if (a1 > b1)
a_score++;
else if (a1 < b1)
b_score++;
else {}
//no op

if (a2 > b2)
a_score++;
else if (a2 < b2)
b_score++;
else {}
//no op

printf("%d %d",a_score, b_score);
return 0;
}``````

### Compare the Triplets Hacker Rank Solution Using JavaScript

``````process.stdin.resume();
process.stdin.setEncoding('ascii');

var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;

process.stdin.on('data', function (data) {
input_stdin += data;
});

process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("\n");
main();
});

return input_stdin_array[input_currentline++];
}

/////////////// ignore above this line ////////////////////

function main() {
var a0 = parseInt(a0_temp);
var a1 = parseInt(a0_temp);
var a2 = parseInt(a0_temp);
var b0 = parseInt(b0_temp);
var b1 = parseInt(b0_temp);
var b2 = parseInt(b0_temp);

let bScore = 0;
let aScore = 0;

for(let i = 0; i < a0_temp.length; i++){
var aProblem = parseInt(a0_temp[i]);
var bProblem = parseInt(b0_temp[i]);
if (aProblem > bProblem) {
aScore++;
}
else if (bProblem > aProblem) {
bScore++;
}
}

console.log(`\${aScore} \${bScore}`);

}``````
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