---
# System prepended metadata

title: LC 2679. Sum in a Matrix
tags: [medium, leedcode, python]

---

# LC 2679. Sum in a Matrix

### [Problem link](https://leetcode.com/problems/sum-in-a-matrix/)

###### tags: `leedcode` `python` `medium`

You are given a  **0-indexed**  2D integer array <code>nums</code>. Initially, your score is <code>0</code>. Perform the following operations until the matrix becomes empty:

- From each row in the matrix, select the largest number and remove it. In the case of a tie, it does not matter which number is chosen.
- Identify the highest number amongst all those removed in step 1. Add that number to your  **score** .

Return the final  **score** .

 **Example 1:** 

```
Input: nums = [[7,2,1],[6,4,2],[6,5,3],[3,2,1]]
Output: 15
Explanation: In the first operation, we remove 7, 6, 6, and 3. We then add 7 to our score. Next, we remove 2, 4, 5, and 2. We add 5 to our score. Lastly, we remove 1, 2, 3, and 1. We add 3 to our score. Thus, our final score is 7 + 5 + 3 = 15.
```

 **Example 2:** 

```
Input: nums = [[1]]
Output: 1
Explanation: We remove 1 and add it to the answer. We return 1.
```

 **Constraints:** 

- <code>1 <= nums.length <= 300</code>
- <code>1 <= nums[i].length <= 500</code>
- <code>0 <= nums[i][j] <= 10<sup>3</sup></code>


## Solution 1
```python=
class Solution:
    def matrixSum(self, nums: List[List[int]]) -> int:
        res = 0
        
        for num in nums:
            num.sort()

        for a in zip(*nums):
            res += max(a)
        
        return res
        
```

>### Complexity
>n =  nums.length
>m = nums[i].length
>|             | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1  | O(nm)           | O(nm)            |

## Note
x
