Electronics Shop in Algorithm | HackerRank Programming Solutions | HackerRank Problem Solving Solutions in Java [💯Correct]

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In this post, you will find the solution for Electronics Shop in Java-HackerRank Problem. We are providing the correct and tested solutions of coding problems present on HackerRank. If you are not able to solve any problem, then you can take help from our Blog/website.

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Introduction To Algorithm

The word Algorithm means “a process or set of rules to be followed in calculations or other problem-solving operations”. Therefore Algorithm refers to a set of rules/instructions that step-by-step define how a work is to be executed upon in order to get the expected results. 

Advantages of Algorithms:

  • It is easy to understand.
  • Algorithm is a step-wise representation of a solution to a given problem.
  • In Algorithm the problem is broken down into smaller pieces or steps hence, it is easier for the programmer to convert it into an actual program.

Link for the ProblemElectronics Shop– Hacker Rank Solution

Electronics Shop – Hacker Rank Solution


A person wants to determine the most expensive computer keyboard and USB drive that can be purchased with a give budget. Given price lists for keyboards and USB drives and a budget, find the cost to buy them. If it is not possible to buy both items, return -1.


image 66

Function Description

Complete the getMoneySpent function in the editor below.

getMoneySpent has the following parameter(s):

  • int keyboards[n]: the keyboard prices
  • int drives[m]: the drive prices
  • int b: the budget


  • int: the maximum that can be spent, or  if it is not possible to buy both items

Input Format

image 67

Sample Input 0

10 2 3
3 1
5 2 8

Sample Output 0


Explanation 0

image 68
Electronics Shop – Hacker Rank Solution
import java.util.Scanner;

 * @author Techno-RJ
public class ElectronicsShop {
	static int getMoneySpent(int[] keyboards, int[] drives, int s) {
		int max = 0;
		int kb = 0, mo = 0;
		for (int i = 0; i < keyboards.length; i++) {
			for (int j = 0; j < drives.length; j++) {
				int tmp = keyboards[i] + drives[j];
				if (tmp >= max && tmp <= s) {
					kb = i;
					mo = j;
					max = tmp;

		return ((kb == 0 && mo == 0) ? -1 : max);


	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int s = in.nextInt();
		int n = in.nextInt();
		int m = in.nextInt();
		int[] keyboards = new int[n];
		for (int keyboards_i = 0; keyboards_i < n; keyboards_i++) {
			keyboards[keyboards_i] = in.nextInt();
		int[] drives = new int[m];
		for (int drives_i = 0; drives_i < m; drives_i++) {
			drives[drives_i] = in.nextInt();
		// The maximum amount of money she can spend on a keyboard and USB drive, or -1
		// if she can't purchase both items
		int moneySpent = getMoneySpent(keyboards, drives, s);

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