1351. Count Negative Numbers in a Sorted Matrix
Easy
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
Constraints:
m == grid.lengthn == grid[i].length1 <= m, n <= 100-100 <= grid[i][j] <= 100
Follow up: Could you find an
O(n + m) solution? class Solution:
def countNegatives(self, grid: List[List[int]]) -> int:
# brute force
count = 0
for col in grid:
for item in col:
if item < 0:
count += 1
return count