# LeetCode 1351. Count Negative Numbers in a Sorted Matrix

## Description

https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/

Given a `m x n` matrix `grid` which is sorted in non-increasing order both row-wise and column-wise, return the number of negative numbers in `grid`.

Example 1:

```Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
Output: 8
Explanation: There are 8 negatives number in the matrix.
```

Example 2:

```Input: grid = [[3,2],[1,0]]
Output: 0
```

Example 3:

```Input: grid = [[1,-1],[-1,-1]]
Output: 3
```

Example 4:

```Input: grid = [[-1]]
Output: 1
```

Constraints:

• `m == grid.length`
• `n == grid[i].length`
• `1 <= m, n <= 100`
• `-100 <= grid[i][j] <= 100`

Follow up: Could you find an `O(n + m)` solution?

## Explanation

Simply count negatives in matrix.

## Python Solution

``````class Solution:
def countNegatives(self, grid: List[List[int]]) -> int:
count = 0

for i in range(len(grid)):
for j in range(len(grid)):
if grid[i][j] < 0:
count += 1

return count
``````
• Time Complexity: O(N^2)
• Space Complexity: O(1)