# 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, b, 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;
}```
```

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

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## Array Manipulation

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