[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:

  1. Use two for loop get value .
  2. If i <0, count=count+1
  3. 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:

https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/discuss/587853/Python-100-space-complexity

https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/discuss/584829/Python-2-lines

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