Fill The Truck/Maximum Units on a Truck Solution Amazon OA 2021

Fill The Truck/Maximum Units on a Truck Solution

You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]:

  • numberOfBoxesi is the number of boxes of type i.
  • numberOfUnitsPerBoxiis the number of units in each box of the type i.

You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize.

See Also : Amazon Online Assessment Questions 2021 Preparation

Return the maximum total number of units that can be put on the truck.

Example 1:

Input: boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
Output: 8
Explanation: There are:
- 1 box of the first type that contains 3 units.
- 2 boxes of the second type that contain 2 units each.
- 3 boxes of the third type that contain 1 unit each.
You can take all the boxes of the first and second types, and one box of the third type.
The total number of units will be = (1 * 3) + (2 * 2) + (1 * 1) = 8.

Example 2:

Input: boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
Output: 91

Constraints:

  • 1 <= boxTypes.length <= 1000
  • 1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000
  • 1 <= truckSize <= 106

Solution

Program C++

bool sortBySecond(const vector<int> &a, const vector<int> &b) {
    
     return (a[1] > b[1]);
}


class Solution 
{
 public:
    int maximumUnits(vector<vector<int>>& boxTypes, int truckSize) {
        
        int total = truckSize;
        int units = 0;
        
        sort(boxTypes.begin(), boxTypes.end(), sortBySecond);
        
        for(auto pp : boxTypes) {
            
            int boxes = pp[0];
            int number_of_units = pp[1];
            
            if(total >= boxes) {
                    units = units + (boxes * number_of_units);
                    total = total - boxes;
                }
            
            else {
                units = units + total*number_of_units;
                total = 0;
            }
        }
        
        return units;
    }
};

Program Python

class Solution:
    def maximumUnits(self, boxTypes: List[List[int]], truckSize: int) -> int:
        maxi = 0
        def sortbasisunits(box):
            return box[1]
		#we are sorting this array in decreasing order w.r.t units of boxes as ultimately we need maximum number of units
        boxTypes.sort(key = sortbasisunits, reverse = True)
        for i in range(len(boxTypes)):
            if boxTypes[i][0] <= truckSize:
                maxi = maxi +  boxTypes[i][0]*boxTypes[i][1]
                truckSize = truckSize - boxTypes[i][0]  #we are calculating the remaining spaces in the truck
            else:
                maxi = maxi + truckSize*boxTypes[i][1]
                truckSize = 0
        return maxi

Program Java

class Solution {
    public int maximumUnits(int[][] boxTypes, int truckSize) {
        Arrays.sort(boxTypes,(a,b)->(b[1]-a[1]));
        
        int i=0;
        int ans=0;
        while(i<boxTypes.length && truckSize>0){
            if(truckSize-boxTypes[i][0]>=0){
                truckSize-=boxTypes[i][0];
                ans+=boxTypes[i][0]*boxTypes[i][1];}else{
                ans+=boxTypes[i][1]*truckSize;
                truckSize=0;
        }
            System.out.println(ans);
            i++;
        }
        return ans;
    }
}

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