**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. Example 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 Returns 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. Constraints 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 :
#!/bin/python3
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++){
if(keyboards[i]+pendrives[j]<=s){
if(max<keyboards[i]+pendrives[j]){
max=keyboards[i]+pendrives[j];
}
}
}
}
System.out.println(max==0?-1:max);
}
}
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++){
scanf("%d",&keyboards[keyboards_i]);
}
int *pendrives = malloc(sizeof(int) * m);
for(int pendrives_i = 0; pendrives_i < m; pendrives_i++){
scanf("%d",&pendrives[pendrives_i]);
}
k=-1;
for(i=0;i<n;i++)
{
if(keyboards[i]>=s)
{
continue;
}
for(j=0;j<m;j++)
{
if(keyboards[i]+pendrives[j]<=s&&keyboards[i]+pendrives[j]>k)
{
k=keyboards[i]+pendrives[j];
}
}
}
printf("%d",k);
return 0;
}
```

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