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

Electronics Shop

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2421 Discussions

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  • kepliu
     + 0 comments
        """
    An optimized version. Instead of checking all combinations, we sort both lists in descending order
    and break the inner loop when b - total cannot improve."""
		
def getMoneySpent(keyboards, drives, b):
    keyboards.sort(reverse = True)
    drives.sort(reverse = True)
    max_spent = -1
    diff = b 
    for k in keyboards:
        for d in drives:
            total = k + d
            if b - total >= 0:
                if b - total < diff:
                    diff = b - total
                    max_spent = total
                else :
                    break
    return max_spent

  • greivin_lopez
     + 0 comments
    from itertools import product
def get_money_spent(keyboards, drives, b):
    possible_sums = [ x + y for x,y in list(product(keyboards, drives)) if x + y <= b ]
    return max(possible_sums) if len(possible_sums) > 0 else -1

  • abdullahabid_561
     + 1 comment
    def getMoneySpent(b, keyboards, drives):
    # 
    # Write your code here.
    final = []
    for i in keyboards:
        for j in drives:
            if (i+j) <= b:
                final.append(i+j)
                
    if len(final) == 0:
        return -1
    else:
        return max(final)

  • tongyou1995
     + 0 comments

    Two-pointer solution:

        keyboards.sort()
    drives.sort(reverse=True)
    
    best = -1
    i = j = 0
    
    while i < len(keyboards) and j < len(drives):
        total = keyboards[i] + drives[j]
        if total > b:
            j += 1
        else:
            if total > best:
                best = total
            i += 1
    return best

  • hristinaglavche1
     + 0 comments

    C# static int getMoneySpent(int[] keyboards, int[] drives, int b) { int max = -1; Array.Sort(keyboards); Array.Sort(drives);

     int i = 0;
 int j = drives.Length - 1;

 while (i <= keyboards.Length - 1 && j >= 0)
 {
     int sum = keyboards[i] + drives[j];

     if (sum > b)
     {
         j--;
     }
     else
     {
         i++;

         if (sum > max)
         {
             max = sum;
         }
     }
 }

 return max;

    }