Published: Sep 13, 2022
This problem might look like dynamic programming. However, the input array’s order does not matter. Sorting and greedy approach works.
You are assigned to put some amount of boxes onto one truck. You are given a 2D array
boxTypes[i] = [numberOfBoxes(i), numberOfUnitsPerBox(i)]:
numberOfBoxes(i)is the number of boxes of type
numberOfUnitsPerBox(i)is the number of units in each box of the type
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. Return the maximum total number of units that can be put on the truck.
1 <= boxTypes.length <= 1000
1 <= numberOfBoxes(i), numberOfUnitsPerBox(i) <= 1000
1 <= truckSize <= 10**6
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
For the first step, sort
boxTypes array by units.
The second step is to go over the sorted
Possible number of boxes to put on the truck is minimum of boxes and current truckSize.
Add up possible number of boxes times current units.
If the truckSize becomes zero, a loop is over.
class MaximumUnitsOnATruck: def maximumUnits(self, boxTypes: List[List[int]], truckSize: int) -> int: boxTypes.sort(key=lambda x: x, reverse=True) amount = 0 for b, u in boxTypes: cur = min(b, truckSize) amount += cur * u truckSize -= cur if truckSize == 0: break return amount