1710. Maximum Units on a Truck (30/06/2022)

1710. Maximum Units on a Truck (30/06/2022)

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Sandeep Chauhan
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Table of contents

Problem:

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.

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 <= numberOfBoxes, numberOfUnitsPerBox <= 1000
  • 1 <= truckSize <= 106

Solution:

Sorting Array:

Algorithm:

  • Sort array boxTypes such that boxType with max units per box are first in the array.
  • Start consuming boxes from starting of the array. Reduce truckSize according to the boxes consumed and add units to result maxUnits.
class Solution {
    public int maximumUnits(int[][] boxTypes, int truckSize) {
        Arrays.sort(boxTypes, (a, b) -> b[1] - a[1]);

        int maxUnits = 0;
        int index = 0;
        while(truckSize > 0 && index < boxTypes.length) {
            int boxes = boxTypes[index][0];
            int unitsPerBox = boxTypes[index][1];

            if(boxes > truckSize) {
                boxes = truckSize;
            }
            maxUnits += (boxes * unitsPerBox);

            truckSize -= boxes;
            index++;
        }

        return maxUnits;
    }
}

Time Complexity: O(n log n) + O(n) where n is length of array. This is required for sorting and iterating through array
Space Complexity: O(1) as we don't need any extra data structure

Creating boxes bucket:

Algorithm:

  • As constraint is provided 1 <= boxTypes.length <= 1000 we can create bucket to store numberOfBoxes on unitPerBox.
  • Start iterating from end of the bucket so we take more number of units per box.
  • On each index reduce the number of units from truckSize and add total units.
class Solution {
    public int maximumUnits(int[][] boxTypes, int truckSize) {
        int[] bucket = new int[1001];
        int maxUnits = 0;

        for(int[] boxType : boxTypes) {
            int boxes = boxType[0];
            int unitPerBox = boxType[1];
            bucket[unitPerBox] += boxes;
        }
        for(int i = 1000; i >= 0 && truckSize > 0; i--) {
            if(bucket[i] == 0) continue;

            int units = Math.min(truckSize, bucket[i]);
            truckSize -= units;
            maxUnits += (units * i);
        }
        return maxUnits;
    }
}

Time Complexity: O(n) + O(1001) = O(n) Where n is length of array. we need to iterate through array boxTypes once and then we need to iterate over bucket of constant size(hence ignored)
Space Complexity: O(1001) = O(1) as we need constant extra space for array bucket

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