Problem Description

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

• numberOfBoxes(i) is the number of boxes of type i.
• numberOfUnitsPerBox(i) is 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. Return the maximum total number of units that can be put on the truck.

Constraints:

• 1 <= boxTypes.length <= 1000
• 1 <= numberOfBoxes(i), numberOfUnitsPerBox(i) <= 1000
• 1 <= truckSize <= 10**6

Examples

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

How to Solve

This problem might look like dynamic programming. However, the input array’s order does not matter. Sorting and greedy approach works.

For the first step, sort box types array by units. The second step is to go over the sorted box types array. 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.

Solution

• C++
• Java
• JavaScript
• Python
• Ruby

• /**
   * @param {number[][]} boxTypes
   * @param {number} truckSize
   * @return {number}
   */
  var maximumUnits = function(boxTypes, truckSize) {
    const bt = boxTypes.sort((a, b) =>  a[1] === b[1] ? b[0] - a[0] : b[1] - a[1])
    let result = 0
    bt.forEach(([b, u]) => {
      let cur = Math.min(b, truckSize)
      result += cur * u
      truckSize -= cur
      if (truckSize === 0) return
    })
    return result
  }
  

• class MaximumUnitsOnATruck:
      def maximumUnits(self, boxTypes: List[List[int]], truckSize: int) -> int:
          boxTypes.sort(key=lambda x: x[1], reverse=True)
          amount = 0
          for b, u in boxTypes:
              cur = min(b, truckSize)
              amount += cur * u
              truckSize -= cur
              if truckSize == 0:
                  break
          return amou
  

• # @param {Integer[][]} box_types
  # @param {Integer} truck_size
  # @return {Integer}
  def maximum_units(box_types, truck_size)
    box_types.sort_by! {|(a, b)| -b}
    result = 0
    box_types.each do |(b, u)|
      cur = [b, truck_size].min
      result += cur * u
      truck_size -= cur
      if truck_size == 0
        break
      end
    end
    result
  end
  

Complexities

• Time: O(nlong(n))
• Space: O(1)