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

Electronics Shop

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2397 Discussions

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  • shrutimadan649
     + 0 comments
            /*
         * Write your code here.
         */       
        Arrays.sort(keyboards);
        Arrays.sort(drives);
        int res=0;
        for(int i=keyboards.length-1;i>=0;i--){
            System.out.println(keyboards[i]);
            if((b-keyboards[i])>0){
                int dr=getDrive(keyboards[i],drives,b);
                System.out.println(dr+" "+keyboards[i]);
                if(dr>0 && (keyboards[i]+dr)>res)
                     res=(keyboards[i]+dr);
            }
        }
        if(res>0)
            return res;
        return -1;
    }
    static int getDrive(int keyp,int[] d,int b){
        for(int j=d.length-1;j>=0;j--){
            if(keyp+d[j]<=b)
            {
                return d[j];
            }
        }
        return -1;
    }

  • patrickgao2020
     + 0 comments

    Here is my Python code! We first find all possible combinations of keyboards and drives and then find the largest sum of those. I appended -1 so if there is no combination within budget, it returns -1.

    def getMoneySpent(keyboards, drives, b):
    possible = list(itertools.product(*[keyboards, drives]))
    for a in range(len(possible)):
        possible[a] = sum(possible[a])
    inbudget = [x for x in possible if x <= b]
    inbudget.append(-1)
    return max(inbudget)

  • Rohanandmail
     + 1 comment

    What is the best time complexity that can be achieved?

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can find the video here : https://youtu.be/yC-TXToDbD0

    int getMoneySpent(vector<int> keyboards, vector<int> drives, int b) {
    int ans = -1;
    for(int k : keyboards){
        for(int d : drives){
            if(k + d > ans && k + d <= b) ans = k + d;
        }
    }
    return ans;
}

  • rabearimananaha1
     + 0 comments

    Very simple python solution here

    def getMoneySpent(keyboards, drives, b):
    sortedKeyBoards = sorted(keyboards, reverse=True)
    drives = sorted(drives, reverse=True)
    sumMax = 0
    
    for i in range(len(sortedKeyBoards)):
        for j in range(len(drives)):
            sumTemp = sortedKeyBoards[i] + drives[j]
            if ( sumTemp >= sumMax and sumTemp <= b):
                sumMax = sumTemp
                
    return -1 if sumMax == 0 else sumMax