# C++ Solution to Coding Challenge 1710. Maximum Units on a Truck

## An example of the greedy algorithm

### Problem statement

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
```

#### Constraints

`1 <= boxTypes.length <= 1000`

.`1 <= numberOfBoxes_i, numberOfUnitsPerBox_i <= 1000`

.`1 <= truckSize <= 10^6`

.

### Solution: Greedy algorithm

#### Code

```
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int maximumUnits(vector<vector<int>>& boxTypes, int truckSize) {
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) {
maxUnits += boxTypes[i][0] * boxTypes[i][1];
truckSize -= boxTypes[i][0];
} else {
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
```

#### Complexity

Runtime:

`O(NlogN)`

, where`N = boxTypes.length`

.Extra space:

`O(1)`

.

### References

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