*This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful,* *please like**this post and/or* *upvote**my solution post on Leetcode's forums.*

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Leetcode Problem #1710 (*Easy*): Maximum Units on a Truck

####
*Description:*

*Description:*

(*Jump to*: *Solution Idea* || *Code*: *JavaScript* | *Python* | *Java* | *C++*)

You are assigned to put some amount of boxes onto

one truck. You are given a 2D array`boxTypes`

, where`boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]`

:

`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 themaximumnumber ofboxesthat 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.maximumtotal number ofunitsthat can be put on the truck

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*Examples:*

*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

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*Constraints:*

*Constraints:*

`1 <= boxTypes.length <= 1000`

`1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000`

`1 <= truckSize <= 10^6`

####
*Idea:*

*Idea:*

(*Jump to*: *Problem Description* || *Code*: *JavaScript* | *Python* | *Java* | *C++*)

For this problem, we simply need to prioritize the more valuable boxes first. To do this, we should **sort** the boxtypes array (**B**) in descending order by the number of units per box (**B[i][1]**).

Then we can iterate through **B** and at each step, we should add as many of the **boxes** as we can, until we reach the truck size (**T**). We should add the number of **boxes** added multiplied by the units per box to our answer (**ans**), and decrease **T** by the same number of **boxes**.

Once the truck is full (**T == 0**), or once the iteration is done, we should **return ans**.

**Time Complexity: O(N log N)**where**N**is the length of**B**, for the sort**Space Complexity: O(1)**

####
*Javascript Code:*

*Javascript Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
var maximumUnits = function(B, T) {
B.sort((a,b) => b[1] - a[1])
let ans = 0
for (let i = 0; T && i < B.length; i++) {
let count = Math.min(B[i][0], T)
ans += count * B[i][1], T -= count
}
return ans
};
```

####
*Python Code:*

*Python Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
class Solution:
def maximumUnits(self, B: List[List[int]], T: int) -> int:
B.sort(key=lambda x: x[1], reverse=True)
ans = 0
for b,n in B:
boxes = min(b, T)
ans += boxes * n
T -= boxes
if T == 0: return ans
return ans
```

####
*Java Code:*

*Java Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
class Solution {
public int maximumUnits(int[][] B, int T) {
Arrays.sort(B, (a,b) -> b[1] - a[1]);
int ans = 0;
for (int[] b : B) {
int count = Math.min(b[0], T);
ans += count * b[1];
T -= count;
if (T == 0) return ans;
}
return ans;
}
}
```

####
*C++ Code:*

*C++ Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
class Solution {
public:
int maximumUnits(vector<vector<int>>& B, int T) {
sort(B.begin(), B.end(), [](auto& a, auto& b) { return b[1] < a[1];});
int ans = 0;
for (auto& b : B) {
int count = min(b[0], T);
ans += count * b[1], T -= count;
if (!T) return ans;
}
return ans;
}
};
```

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