# [LeetCode]#1351. Count Negative Numbers in a Sorted Matrix

**Environment: Python 3.7**

**Key technique:** for

Given a `m * 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`

**Analysis:**

- Use two for loop get value .
- If i <0, count=count+1
- return count

**Solution:**

`class Solution:`

def countNegatives(self, grid):

count=0

for i in grid:

for j in i:

if j<0:

count+=1

return count

**Better Solution:**

`class Solution:`

def countNegatives(self, grid):

i = [item for row in grid for item in row]

return len([j for j in i if j<0])

**Submitted result:**

**Reference:**