Maximum Units on a Truck

Published: Sep 13, 2022

Easy Sorting Greedy

Introduction

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

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

https://leetcode.com/problems/maximum-units-on-a-truck/

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

Analysis

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

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 amount

Complexities

  • Time: O(nlong(n))
  • Space: O(1)