**Compare the Triplets**

### Problem Statement :

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. Example 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. Function Description Complete the function compareTriplets in the editor below. compareTriplets has the following parameter(s): int a[3]: Alice's challenge rating int b[3]: Bob's challenge rating Return int[2]: 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[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

### Solution :

` ````
Solution in C :
In C++ :
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define REP(i, n) for (int i = 0; i < (int)(n); ++i)
typedef long long LL;
typedef pair<int, int> PII;
int a[3], b[3];
int main() {
REP(i, 3) scanf("%d", a + i);
REP(i, 3) scanf("%d", b + i);
int x = 0, y = 0;
REP(i, 3) if (a[i] > b[i]) {
++x;
} else if (a[i] < b[i]) {
++y;
}
printf("%d %d\n", x, y);
return 0;
}
In C :
int* compareTriplets(int a_count, int* a, int b_count, int* b, int* result_count) {
int Alice_point = 0, Bob_point = 0;
for(int i = 0; i<a_count; i++)
{
if(a[i]==b[i])
{
}
if(a[i]>b[i])
{
Alice_point += 1;
}
if(a[i]<b[i])
{
Bob_point += 1;
}
}
*result_count = 2;
static int c[2];
c[0] = Alice_point;
c[1] = Bob_point;
return c;
}
In Java:
import java.io.*;
import java.util.*;
public class Solution {
private BufferedReader in;
private StringTokenizer line;
private PrintWriter out;
private static final int mm = 1000000007;
public void solve() throws IOException {
int[] a = nextIntArray(3);
int[] b = nextIntArray(3);
int aa = 0;
int bb = 0;
for (int i = 0; i < 3; i++) {
if (a[i] > b[i]) aa++;
else if (a[i] < b[i]) bb++;
}
out.println(aa + " " + bb);
}
public static void main(String[] args) throws IOException {
new Solution().run(args);
}
public void run(String[] args) throws IOException {
if (args.length > 0 && "DEBUG_MODE".equals(args[0])) {
in = new BufferedReader(new InputStreamReader(new FileInputStream("input.txt")));
} else {
in = new BufferedReader(new InputStreamReader(System.in));
}
out = new PrintWriter(System.out);
// out = new PrintWriter("output.txt");
// int t = nextInt();
int t = 1;
for (int i = 0; i < t; i++) {
// out.print("Case #" + (i + 1) + ": ");
solve();
}
in.close();
out.flush();
out.close();
}
private int[] nextIntArray(int n) throws IOException {
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = nextInt();
}
return res;
}
private long[] nextLongArray(int n) throws IOException {
long[] res = new long[n];
for (int i = 0; i < n; i++) {
res[i] = nextInt();
}
return res;
}
private int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
private long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
private double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
private String nextToken() throws IOException {
while (line == null || !line.hasMoreTokens()) {
line = new StringTokenizer(in.readLine());
}
return line.nextToken();
}
private static class Pii {
private int key;
private int value;
public Pii(int key, int value) {
this.key = key;
this.value = value;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pii pii = (Pii) o;
if (key != pii.key) return false;
return value == pii.value;
}
@Override
public int hashCode() {
int result = key;
result = 31 * result + value;
return result;
}
@Override
public String toString() {
return "Pii{" +
"key=" + key +
", value=" + value +
'}';
}
}
private static class Pair<K, V> {
private K key;
private V value;
public Pair(K key, V value) {
this.key = key;
this.value = value;
}
public K getKey() {
return key;
}
public V getValue() {
return value;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
if (key != null ? !key.equals(pair.key) : pair.key != null) return false;
return !(value != null ? !value.equals(pair.value) : pair.value != null);
}
@Override
public int hashCode() {
int result = key != null ? key.hashCode() : 0;
result = 31 * result + (value != null ? value.hashCode() : 0);
return result;
}
}
}
In Python3 :
#!/bin/python3
import sys
a0,a1,a2 = input().strip().split(' ')
a0,a1,a2 = [int(a0),int(a1),int(a2)]
b0,b1,b2 = input().strip().split(' ')
b0,b1,b2 = [int(b0),int(b1),int(b2)]
A = (a0>b0) + (a1>b1) + (a2>b2)
B = (a0<b0) + (a1<b1) + (a2<b2)
print(A, B)
```

## View More Similar Problems

## Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →## The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

View Solution →## Self-Driving Bus

Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever

View Solution →## Unique Colors

You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti

View Solution →## Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

View Solution →## Pair Sums

Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v

View Solution →