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 .

get 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 . from screen.

Electronics Shop

Determine the most expensive Keyboard and USB drive combination one can purchase within her budget.

Electronics Shop Problem Submissions Leaderboard Editorial

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 .

Example

The person can buy a , or a . Choose the latter as the more expensive option and return .

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 if it is not possible to buy both items

Input Format

The first line contains three space-separated integers , , and , the budget, the number of keyboard models and the number of USB drive models.

The second line contains space-separated integers , the prices of each keyboard model.

The third line contains space-separated integers , the prices of the USB drives.

Constraints

The price of each item is in the inclusive range .

Sample Input 0

10 2 3 3 1 5 2 8

Sample Output 0

9

Explanation 0

Buy the keyboard and the USB drive for a total cost of .

Sample Input 1

5 1 1 4 5

Sample Output 1

-1

Explanation 1

There is no way to buy one keyboard and one USB drive because , so return .

vector

Change Theme C++ 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 fout.close(); return 0; }

 split_string(string input_string) {

string::iterator new_end = unique

(input_string.begin(), input_string.end(), []

(const char &x, const char &y) {

return x == y and x == ' ';

});

input_string.erase(new_end, input_string.end()

);

while (input_string[input_string.length() - 1]

== ' ') {

input_string.pop_back();

vector

}

 splits;

char delimiter = ' ';

size_t i = 0;

size_t pos = input_string.find(delimiter);

while (pos != string::npos) {

splits.push_back(input_string.substr(i,

pos - i)); i = pos + 1;

pos = input_string.find(delimiter, i);

}

splits.push_back(input_string.substr(i, min

(pos, input_string.length()) - i + 1));

return splits; } Line: 99 Col: 1

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Electronics Shop

A person wants to determine the most expensive computer keyboard and USB drive that can be purchased with a give …

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();

for(int i=0;i

} int max=0;

for(int j=0;j

if(keyboards[i]+pendrives[j]<=s){

if(max

max=keyboards[i]+pendrives[j];

} } } }

System.out.println(max==0?-1:max);

#include

} } C++ :

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);

#include

return 0; } C :

#include #include #include #include #include #include

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]);

for(i=0;i

} k=-1;

for(j=0;j

{ if(keyboards[i]>=s) { continue; }

{

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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स्रोत : hackerranksolution.in

Electronics Shop

Hello coders, today we are going to solve Electronics Shop HackerRank Solution which is a Part of HackerRank Algorithm Series.

Electronics Shop | HackerRank Solution

Leave a Comment / HackerRank, HackerRank Algorithms / By CodeBros

Hello coders, today we are going to solve Electronics Shop HackerRank Solution which is a Part of HackerRank Algorithm Series.

Task

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.

Exampleb = 60keyboards = [40, 50, 60]drivers = [5, 8, 12]

The person can buy a 40 keyboard + 12 USB drive = 52, or a 50 keyboard + 8 USB drive = 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 < 10001 <= b <= 106

The price of each item is in the inclusive range [1, 106].

Sample Input 0

10 2 3 3 1 5 2 8

Sample Output 0

9

Explanation 0

Buy the 2nd keyboard and the 3rd USB drive for a total cost of 8 + 1 = 9.

Sample Input 1

5 1 1 4 5

Sample Output 1

-1

Explanation 1

There is no way to buy one keyboard and one USB drive because 4 + 5 > 5, so return -1.

Solution – Electronics Shop

C++

#include

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; }

Python

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();

for(int i=0;i

} int max=0;

for(int j=0;j

if(keyboards[i]+pendrives[j]<=s){

if(max

max=keyboards[i]+pendrives[j];

} } } }

System.out.println(max==0?-1:max);

} }

Disclaimer: The above Problem (Electronics Shop) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.

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