1081MediumArrayDepth-First SearchBreadth-First SearchUnion-FindMatrixCounting

Which Units Can Reach Another

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Trains the technique from

LeetCode 1267Count Servers that Communicate

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A rack is laid out as the grid grid, where 1 marks a cell holding a unit and 0 marks an empty cell.

Two units can reach each other when they share a row or share a column.

Return how many units can reach at least one other unit.

Examples

Example 1

Input: grid = [[1, 1, 1]]
Output: 3

All three units share the single row, so each reaches the other two.

Example 2

Input: grid = [[1, 0, 0], [0, 1, 0], [0, 0, 1]]
Output: 0

Each unit has its own row and its own column to itself, so none can reach another.

Example 3

Input: grid = [[0, 1, 0], [1, 0, 1], [0, 1, 0]]
Output: 4

The two units in the middle row reach each other, and the units above and below share their column with the middle one, so all four count.

Constraints

1 <= grid.length <= 250
1 <= grid[i].length <= 250
0 <= grid[i][j] <= 1

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_servers(grid: list[list[int]]) -> int:

Java

public int countServers(int[][] grid)

The intended approach

Time O(m * n) over the grid's cells.Space O(m + n) for the tallies.

A unit can reach another exactly when its own row holds more than one unit, or its own column does. Nothing further than that matters, since reaching is decided by sharing a row or a column and not by any chain of hops. So tally the units in each row and in each column in one sweep of the grid. Then sweep again and count every unit whose row tally or column tally is above one. Two sweeps and two lists of tallies.

Pythonentry point: count_servers

def count_servers(grid):
    rows = len(grid)
    cols = len(grid[0])
    row_tally = [0] * rows
    col_tally = [0] * cols
    for r in range(rows):
        for c in range(cols):
            if grid[r][c]:
                row_tally[r] += 1
                col_tally[c] += 1
    total = 0
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] and (row_tally[r] > 1 or col_tally[c] > 1):
                total += 1
    return total

The obvious version

Test each unit against every other unit outright, counting the unit as soon as one of them shares its row or its column. Costs the units squared rather than two sweeps with tallies.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_servers(grid):
    rows = len(grid)
    cols = len(grid[0])
    spots = []
    for r in range(rows):
        for c in range(cols):
            if grid[r][c]:
                spots.append((r, c))
    total = 0
    for i in range(len(spots)):
        for j in range(len(spots)):
            if i == j:
                continue
            if spots[i][0] == spots[j][0] or spots[i][1] == spots[j][1]:
                total += 1
                break
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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