How do you solve Count Servers that Communicate on LeetCode?

Count Servers that Communicate (LeetCode #1267) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m + n) space complexity. Key insight: Servers communicate if they share the same row or column.

What is the brute force solution for Count Servers that Communicate?

The brute force approach for Count Servers that Communicate is Brute Force, with O(m * n * (m + n)) time complexity and O(1) space complexity. In this approach, we will check each server and see if it can communicate with any other server by scanning its entire row and column. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(m * n).

What are the interview tips for Count Servers that Communicate?

Always clarify the problem and constraints before jumping to code. Think about how to optimize your approach early on. Practice explaining your thought process as you code.

What are common mistakes when solving Count Servers that Communicate?

Not considering edge cases where servers are isolated. Forgetting to check both row and column counts.

#1267

Count Servers that Communicate

Medium

ArrayDepth-First SearchBreadth-First SearchUnion-FindMatrixCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * (m + n))	O(m * n)
Space	O(1)	O(m + n)

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Intuition

Time O(m * n)Space O(m + n)

We can optimize the process by counting the number of servers in each row and column first. This allows us to determine if a server can communicate without scanning the entire grid multiple times.

⚙️

Algorithm

4 steps

1Step 1: Initialize two arrays to count servers in each row and each column.
2Step 2: Traverse the grid and populate these counts.
3Step 3: Traverse the grid again, and for each server, check if its row or column count is greater than 1.
4Step 4: If yes, increment the total count of communicative servers.

solution.py14 lines

1def countServers(grid):
2    row_count = [0] * len(grid)
3    col_count = [0] * len(grid[0])
4    for i in range(len(grid)):
5        for j in range(len(grid[0])):
6            if grid[i][j] == 1:
7                row_count[i] += 1
8                col_count[j] += 1
9    count = 0
10    for i in range(len(grid)):
11        for j in range(len(grid[0])):
12            if grid[i][j] == 1 and (row_count[i] > 1 or col_count[j] > 1):
13                count += 1
14    return count

ℹ

Complexity note: This complexity is efficient because we only traverse the grid a couple of times, leading to linear time complexity relative to the number of cells.

1Servers communicate if they share the same row or column.
2Counting servers in rows and columns can simplify the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.