Published: Sep 13, 2022

## 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`

## 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)`