Count Servers that Communicate Python Solution

Tested Python solution for LeetCode 1267 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 1267, Medium. Topics: Array, Depth-First Search, Breadth-First Search, Union-Find, Matrix, Counting. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 18 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1267   # by problem number
lcpy gen -s count_servers_that_communicate   # by problem name

​

Problem

You are given a map of a server center, represented as a m * n integer matrix grid, where 1 means that on that cell there is a server and 0 means that it is no server. Two servers are said to communicate if they are on the same row or on the same column.

Return the number of servers that communicate with any other server.

​

Examples

Input: grid = [[1,0],[0,1]]
Output: 0
Explanation: No servers can communicate with others.

Input: grid = [[1,0],[1,1]]
Output: 3
Explanation: All three servers can communicate with at least one other server.

Input: grid = [[1,1,0,0],[0,0,1,0],[0,0,1,0],[0,0,0,1]]
Output: 4
Explanation: The two servers in the first row can communicate with each other. The two servers in the third column can communicate with each other. The server at right bottom corner can't communicate with any other server.

​

Constraints

• m == grid.length
• n == grid[i].length
• 1 <= m <= 250
• 1 <= n <= 250
• grid[i][j] == 0 or 1

​

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    def count_servers(self, grid: list[list[int]]) -> int:
        m, n = len(grid), len(grid[0])
        rows = [sum(row) for row in grid]
        cols = [sum(grid[i][j] for i in range(m)) for j in range(n)]
        return sum(
            grid[i][j] == 1 and (rows[i] > 1 or cols[j] > 1) for i in range(m) for j in range(n)
        )

​

Complexity

Time	Space
-	-

​

Tags

NeetCode All.