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[0])):
if grid[i][j] < 0:
count += 1
return count
- Time Complexity: O(N^2)
- Space Complexity: O(1)