Maximum Units on a Truck#

Problem statement#

[1]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.

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
Explanation: (5 * 10) + (3 * 9) + (2 * 7) = 91.

Constraints#

  • 1 <= boxTypes.length <= 1000.

  • 1 <= numberOfBoxes_i, numberOfUnitsPerBox_i <= 1000.

  • 1 <= truckSize <= 10^6.

Solution: Greedy algorithm#

Put the boxes having more units first.

Code#

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int maximumUnits(vector<vector<int>>& boxTypes, int truckSize) {
    // sort for the boxes based on their number of units
    sort(boxTypes.begin(), boxTypes.end(), 
         [](const vector<int>& a, const vector<int>& b) {
            return a[1] > b[1];
         });
    int maxUnits = 0;
    int i = 0;
    while (truckSize > 0 && i < boxTypes.size()) {
        if (boxTypes[i][0] <= truckSize) {
            // put all boxTypes[i] if there is still room
            maxUnits += boxTypes[i][0] * boxTypes[i][1];
            truckSize -= boxTypes[i][0];
        } else {
            // can put only truckSize < boxTypes[i][0] of boxTypes[i]
            maxUnits += truckSize * boxTypes[i][1];
            break;
        }
        i++;
    }
    return maxUnits;
}
int main() {
    vector<vector<int>> boxTypes{{1,3},{2,2},{3,1}};
    cout << maximumUnits(boxTypes, 4) << endl;
    boxTypes = {{5,10},{2,5},{4,7},{3,9}};
    cout << maximumUnits(boxTypes, 10) << endl;
}

Output:
8
91

This solution optimally loads boxes onto a truck to maximize the total number of units that can be transported, considering both the number of boxes available and their units per box.

Complexity#

  • Runtime: O(N*logN), where N = boxTypes.length.

  • Extra space: O(1).

Modern C++ STL notes#

Note that two vectors[2] can be compared. That is why you can sort them.

But in this case you want to sort them based on the number of units. That is why you need to define the comparison function like the code above. Otherwise, the std::sort[3] algorithm will use the dictionary order to sort them by default.

Exercise#

  • Maximum Bags With Full Capacity of Rocks[4].