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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Electronics Shop
  5. Discussions

Electronics Shop

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  • ryanfehr18
     + 86 comments

    Java Solution Here is my solution in Java that passes all cases in O(n log(n)) time with O(1) additional space.

    Here is how it works: We sort keyboards in descending order. We sort usb in ascending order. Then we iterate over them not checking usbs past where 1 usb plus our current keyboard is already greater than s. Read the comments in my solution for a more in depth explanation.

    Some times HackeRank redirect doesn't work, so here is the direct link you will have to copy and paste: https://github.com/RyanFehr/HackerRank/blob/master/Algorithms/Implementation/Electronics%20Shop/Solution.java

  • SergZerg
     + 11 comments

    My python 3 solution: 

    def getMoneySpent(keyboards, drives, b):
    l = []
    for x in keyboards:
        for y in drives:
            if x+y <= b:
                l.append(x+y)
    if not l:
        return -1
    else:
        return max(l)

  • LeHarkunwar
     + 9 comments

    Well, did it the pythonic 1 line way

    def getMoneySpent(keyboards, drives, s):
        return max([sum([x,y]) for x in keyboards for y in drives if sum([x,y]) <= s]+[-1])

  • Himanshu_1061
     + 19 comments
    //Try This One...
//Coded By Himanshu And Jatin....
//Chandigarh University....

#include <bits/stdc++.h>

using namespace std;


int main() {
    int s;
    int n;
    int m;
    cin >> s >> n >> m;
    int k[n]; 
    int z=-1;
    int u[m];
    
    for(int i=0;i<n;i++)
    { cin >> k[i];
    }
    for(int i=0;i<m;i++)
    {
     cin >> u[i];    } 
    for(int i=0;i<n;i++)
    {
    for(int j=0;j<m;j++)
        {
        if(k[i]+u[j]<=s && k[i]+u[j]>z)
            z=k[i]+u[j];
    }
}
    cout<<z;
    return 0;
}

  • mbournehalley
     + 3 comments

    Javascript solution:

    function getMoneySpent(keyboards, drives, b) {
  return keyboards.reduce((acc, curr) =>
    Math.max(acc, ...drives.map(usb => usb + curr).filter(ku => b >= ku))
  , -1);
}
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