Electronics Shop

Problem Statement :

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.

b = 60
keyboards = [40, 50, 60]
drives = [5, 8,12]

The person can buy a 40 keyboards + 12 USB drives = 52, or a 50 keyboards + 8 USB drives = 58. Choose the latter as the more expensive option and return 58.

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 -1 if it is not possible to buy both items

Input Format

The first line contains three space-separated integers b, n, and m, the budget, the number of keyboard models and the number of USB drive models.
The second line contains n space-separated integers keyboard[i], the prices of each keyboard model.
The third line contains m space-separated integers drives, the prices of the USB drives.

1 <= n,m <= 1000
1 <= b <= 10^6
The price of each item is in the inclusive range [1, 10^6].

Solution :


                            Solution in C :

python 3 :


import sys

s,n,m = input().strip().split(' ')
s,n,m = [int(s),int(n),int(m)]
a = [int(keyboards_temp) for keyboards_temp in input().strip().split(' ')]
b = [int(pendrives_temp) for pendrives_temp in input().strip().split(' ')]
ans = -1
for x in a:
    for y in b:
        if x + y <= s:
            ans = max(ans, x + y)
print (ans)

Java   :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    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[] pendrives = new int[m];
        for(int pendrives_i=0; pendrives_i < m; pendrives_i++){
            pendrives[pendrives_i] = in.nextInt();
        int max=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){

C++  :

#include <bits/stdc++.h>

using namespace std;

int ans=-1, a[1005], b[1005], i, n, m, s, j;

int main()

    scanf("%d%d%d", &s, &n, &m);
    for(i=1; i<=n; ++i) scanf("%d", &a[i]);
    for(i=1; i<=m; ++i) scanf("%d", &b[i]);

    for(i=1; i<=n; ++i)
    for(j=1; j<=m; ++j)
        if(a[i]+b[j]<=s) ans = max(a[i]+b[j], ans);

    printf("%d\n", ans);

    return 0;

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 s; 
    int n; 
    int m,i,j,k,l; 
    scanf("%d %d %d",&s,&n,&m);
    int *keyboards = malloc(sizeof(int) * n);
    for(int keyboards_i = 0; keyboards_i < n; keyboards_i++){
    int *pendrives = malloc(sizeof(int) * m);
    for(int pendrives_i = 0; pendrives_i < m; pendrives_i++){
    return 0;

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