Leetcode 1710. Maximum Units on a Truck

Accepted Code

Problem Statement

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.
  • numberOfUnitsPerBoxi 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.

Test Cases

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


  • 1 <= boxTypes.length <= 1000
  • 1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000
  • 1 <= truckSize <= 10^6


Our task is to find the maximum total number of units that can be put on the truck. We are given a 2d array where boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]. So the idea is quite straight that we just need to do a sort based on numberOfUnitsPerBox in descending order. Then just calculate the unit until our track have enough size.


  1. Sort the given 2d array based on numberOfUnitsPerBox in descending order.
  2. Traverse through the sorted array.
  3. If we have enough truck size, put the box into the truck and calculate the unit.
  4. Stop the iteration if we can’t no more hold anymore box in the truck.
  5. Return the resultant unit which will be the maximum total number of unit.

Time and Space Complexity

Here we are sorting the given array and traversing through the array once. So if the length of array=n, then time complexity will be O(nlogn + n).

We are not making use of any extra space, so space complexity will be O(1).


Java Code



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Rohith Vazhathody

Rohith Vazhathody

Software Engineer | Weekly articles | Interested in DSA, design | Writes about problem solving, algorithm explanation of my understanding and Java Codes.