Electronics Shop

#1

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

Sample Input 1

5 1 1
4
5

Sample Output 1

-1

#2

first sort both the vectors and then run two loops

Reference code

int getMoneySpent(vector<int> keyboards, vector<int> drives, int b) {
    int n=keyboards.size();
    int m=drives.size();
    sort(keyboards.begin(),keyboards.end());
    sort(drives.begin(),drives.end());
    int ans=INT_MIN;
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            int temp=keyboards[i]+drives[j];
            if(temp<=b){
                ans=max(ans,temp);
            }
            else break;
        }
    }
    if(ans==INT_MIN)return -1;
    return ans;
}
#3

thanks but it would be good if u provide cb ide soln as direct text ommits >> into lt etc

#4

this is the code

if your doubt is resolved
don’t forgot to mark it as resolved

#5

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