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

Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

View Solution →

Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

View Solution →

Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

View Solution →

Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

View Solution →

Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →

Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →