Naive Solution

A brute force solution would involve iterating through each cell in the grid. For each cell containing a server (value of 1), we check if there is another server in the same row or column. If there is, we increment a counter. This approach is straightforward but not the most efficient.

Code (Python)

def count_servers_brute_force(grid):
    count = 0
    rows = len(grid)
    cols = len(grid[0])

    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 1:
                communicating = False
                # Check row
                for k in range(cols):
                    if k != j and grid[i][k] == 1:
                        communicating = True
                        break
                # Check column
                if not communicating:
                    for k in range(rows):
                        if k != i and grid[k][j] == 1:
                            communicating = True
                            break
                if communicating:
                    count += 1
    return count

• Time Complexity: O(m * n * (m + n)), where m is the number of rows and n is the number of columns.
• Space Complexity: O(1)

Optimal Solution

A more efficient solution involves pre-calculating the number of servers in each row and each column. Then, we iterate through the grid again. If a cell contains a server, we check if its row count or its column count is greater than 1. If it is, it means that the server can communicate with at least one other server. This approach avoids redundant checks.

Code (Python)

def count_servers_optimal(grid):
    rows = len(grid)
    cols = len(grid[0])

    row_counts = [0] * rows
    col_counts = [0] * cols

    # Calculate row and column counts
    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 1:
                row_counts[i] += 1
                col_counts[j] += 1

    count = 0
    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 1 and (row_counts[i] > 1 or col_counts[j] > 1):
                count += 1

    return count

• Time Complexity: O(m * n), where m is the number of rows and n is the number of columns.
• Space Complexity: O(m + n), due to the extra space used for row_counts and col_counts arrays.

Edge Cases

• Empty grid: Should return 0.
• Grid with no servers: Should return 0.
• Grid with only one server: Should return 0.
• Grid with all servers in one row or column: Should return the number of servers.
• Grid with a mix of communicating and non-communicating servers: Should return the count of communicating servers only.