Problem Description

In this problem, we're given a representation of a server center as a m * n integer matrix called grid, where each cell in the grid can either contain a server (represented by the number 1) or not contain a server (represented by the number 0). The crucial point to understand is that servers are considered to be able to communicate with each other if and only if they are located in the same row or the same column. The objective is to calculate how many servers are able to communicate with at least one other server within the grid.

Flowchart Walkthrough

Let's analyze the problem using the Flowchart. Here's a step-by-step walkthrough for LeetCode 1267. Count Servers that Communicate:

1. Is it a graph?

   • Yes: The grid of servers can be considered a graph where each server (represented by a value of 1) is a node, and there are implied connections along rows and columns if there are multiple servers.

2. Is it a tree?

   • No: There isn't necessarily a hierarchical structuring among servers, and connections don't form parent-child relationships.

3. Is the problem related to directed acyclic graphs (DAG)?

   • No: The problem is about connecting servers through rows and columns, which doesn't involve directed acyclic properties or topological sorting.

4. Is the problem related to shortest paths?

   • No: The goal is to count servers that can communicate, not to find the shortest path between servers.

5. Does the problem involve connectivity?

   • Yes: The problem requires counting the servers in connected components, where each connected component forms by servers communicating either in the same row or column.

6. Is the graph weighted?

   • No: There are no weights involved in the server grid; it's simply about presence (1) or absence (0) of a server at any position.

Based on these answers, we conclude that the problem should be addressed by exploring connectivity in the grid where servers are nodes. Since the grid is unweighted and involves checking connected formations, an appropriate approach to solving this would be either Depth-First Search (DFS) or Breadth-First Search (BFS). Given we can efficiently handle DFS in this context, and since DFS can be utilized recursively to explore all connected servers once a communicating server is found, DFS would be an eligible method to apply. Thus, Depth-First Search pattern is a relevance in solving this problem.

Intuition

To determine the number of servers that can communicate with others, we need a way to check each server and see if there is at least one other server on the same row or column. The solution approach can start by keeping two separate arrays, row and col, to keep track of how many servers are on each row and column respectively.

We then iterate over the entire grid, counting how many servers are in each row and in each column. This is done by going through each cell of the matrix, and whenever we encounter a server (a cell with a value of 1), we increment the corresponding row and column counters by 1.

Once we have the counts of servers in each row and column, we can again iterate over the grid. This time, for each server, we check whether the corresponding row or column has more than one server (which means the count is greater than 1). If so, the server can communicate with others, and it should be included in the final count. We do this check using a generator expression and pass it to sum function to get the total count of servers that can communicate.

This approach efficiently factors in both row-wise and column-wise communication, ensuring no double counting of servers.

Learn more about Depth-First Search, Breadth-First Search and Union Find patterns.

Solution Approach

The implementation of the solution uses a two-pass algorithm across the rows and columns of the server grid matrix. The algorithm selectively counts and then aggregates server connections based on the communication criteria. The solution leverages simple data structures – two lists named row and col – to store the counts of servers in each row and column, respectively.

Here's a step-by-step walk-through of the algorithm:

1. Initialize two lists, row and col, with a size equal to the number of rows m and columns n in input grid, respectively, and set all values to 0. These lists hold counts of servers in each row and column.

   row = [0] * m
   col = [0] * n

2. Iterate over each cell in the grid matrix to fill in the row and col arrays. Whenever a server (1 in grid) is found, increment the respective row's and column's count.

   for i in range(m):
       for j in range(n):
           if grid[i][j]:
               row[i] += 1
               col[j] += 1

3. After populating the row and col arrays, we carry out a second iteration over the grid. For each server found, check if it can communicate (i.e., if row[i] or col[j] is greater than 1).

   The expression grid[i][j] and (row[i] > 1 or col[j] > 1) evaluates to True if there is a server in grid[i][j] and it can communicate with at least one server in the same row or column.

   The generator expression inside the sum function goes through all cells in the grid, evaluates the above condition, and if it is True, it contributes 1 toward the sum, effectively counting the server.

   return sum(
       grid[i][j] and (row[i] > 1 or col[j] > 1)
       for i in range(m)
       for j in range(n)
   )

This code bypasses a separate conditional branching for each server and instead uses a compact generator expression to do the counting, which is a common Pythonic pattern for concise and readable code. The main algorithmic insight is that by keeping track of the row and column server counts, we can determine the capability of each server to communicate in constant time during the second iteration.

Example Walkthrough

Let's assume we have a server grid grid represented by the following 3 x 4 matrix:

[
  [1, 0, 0, 1],
  [0, 1, 1, 0],
  [0, 0, 1, 0]
]

Let's walk through the solution step-by-step for this example:

1. Initialize two lists, row and col, each with values initialized at 0. For our grid, row will have 3 elements and col will have 4 elements, corresponding to the number of rows and columns in grid, respectively.

   After initialization:

   row = [0, 0, 0]
   col = [0, 0, 0, 0]

2. We iterate over each cell in the grid to count the number of servers in each row and column.

   After completing this step for our example, the row and col lists will look like this:

   row = [2, 2, 1]
   col = [1, 1, 2, 1]

   The row list tells us that the first and second rows each have 2 servers, while the third row has 1 server. The col list tells us that the first, second, and fourth columns each have 1 server, and the third column has 2 servers.

3. Next, we perform a second iteration over the grid. This time, for each server, we check if it can communicate with others. If row[i] or col[j] is greater than 1, the server can communicate.

   • For grid[0][0], we have a server and row[0] > 1 (because row[0] is 2) - so this server can communicate.
   • For grid[0][3], we have a server, but neither row[0] > 1 nor col[3] > 1 (because row[0] is 2, but col[3] is 1) - so this server cannot communicate.
   • For grid[1][1], we have a server and both row[1] > 1 and col[1] > 1, which means this server can communicate.
   • For grid[1][2], we have a server and col[2] > 1, so this server can communicate.
   • For grid[2][2], we do have a server but since row[2] is not greater than 1, this server cannot communicate.

   We continue performing this check for every cell in the grid that contains a server (1).

   Using the generator expression provided, we count the servers that can communicate, which evaluates to 3 for our example (grid[0][0], grid[1][1], and grid[1][2]).

Therefore, for the grid given in our example, the total number of servers that can communicate with at least one other server is 3.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code consists of two main parts: The first part is a double loop that goes through the entire grid to count the number of servers in each row and column. Since this loop goes through all the elements of the m x n grid exactly once, the time complexity for this part is O(m * n).

The second part of the code calculates the sum with a generator expression that also iterates through every element in the grid. It checks if there's a server in the given cell (grid[i][j]) and if there is more than one server in the corresponding row or column. Since this is done for each element, it also has a time complexity of O(m * n).

Therefore, the total time complexity of the entire function is O(m * n) + O(m * n), which simplifies to O(m * n) because we only take the highest order term for big O notation.

Space Complexity

For space complexity, the code uses additional arrays row and col to store the counts of servers in each row and column, respectively. The size of row is m and the size of col is n. Hence, the extra space used is O(m + n).

In conclusion, the time complexity is O(m * n) and the space complexity is O(m + n).

Learn more about how to find time and space complexity quickly using problem constraints.