LeetCode #1267 — MEDIUM

Count Servers that Communicate

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Example 1:

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

Example 2:

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

Example 3:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Union-Find

Example 1

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

Example 2

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

Example 3

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

Step 02

Core Insight

What unlocks the optimal approach

Store number of computer in each row and column.
Count all servers that are not isolated.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1267: Count Servers that Communicate
class Solution {
    public int countServers(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[] row = new int[m];
        int[] col = new int[n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                row[i] += grid[i][j];
                col[j] += grid[i][j];
            }
        }
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1 && (row[i] > 1 || col[j] > 1)) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1267: Count Servers that Communicate
func countServers(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	row, col := make([]int, m), make([]int, n)
	for i := range grid {
		for j, x := range grid[i] {
			row[i] += x
			col[j] += x
		}
	}
	for i := range grid {
		for j, x := range grid[i] {
			if x == 1 && (row[i] > 1 || col[j] > 1) {
				ans++
			}
		}
	}
	return
}

# Accepted solution for LeetCode #1267: Count Servers that Communicate
class Solution:
    def countServers(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        row = [0] * m
        col = [0] * n
        for i in range(m):
            for j in range(n):
                row[i] += grid[i][j]
                col[j] += grid[i][j]
        return sum(
            grid[i][j] and (row[i] > 1 or col[j] > 1)
            for i in range(m)
            for j in range(n)
        )

// Accepted solution for LeetCode #1267: Count Servers that Communicate
impl Solution {
    pub fn count_servers(grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let mut row = vec![0; m];
        let mut col = vec![0; n];
        for i in 0..m {
            for j in 0..n {
                row[i] += grid[i][j];
                col[j] += grid[i][j];
            }
        }
        let mut ans = 0;
        for i in 0..m {
            for j in 0..n {
                if grid[i][j] == 1 && (row[i] > 1 || col[j] > 1) {
                    ans += 1;
                }
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #1267: Count Servers that Communicate
function countServers(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const row = new Array(m).fill(0);
    const col = new Array(n).fill(0);
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            row[i] += grid[i][j];
            col[j] += grid[i][j];
        }
    }
    let ans = 0;
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (grid[i][j] === 1 && (row[i] > 1 || col[j] > 1)) {
                ans++;
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m + n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.