0958MediumArrayHash TableEnumeration

Marked Cells in Every Two-by-Two Window

Tracked in this browser only

Trains the technique from

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 sheet has m rows and n columns of cells, numbered from zero. The cells listed in coordinates, each [row, column], are marked; the rest are blank, and no cell is listed twice.

A window is any two-by-two square of neighbouring cells, so there are (m - 1) * (n - 1) of them.

Return a list of five counts: how many windows hold exactly 0 marked cells, then exactly 1, and so on up to exactly 4.

Examples

Example 1

Input: m = 2, n = 2, coordinates = [[0, 0], [0, 1], [1, 0], [1, 1]]
Output: [0, 0, 0, 0, 1]

The sheet is a single window and all four of its cells are marked.

Example 2

Input: m = 3, n = 3, coordinates = [[1, 1]]
Output: [0, 4, 0, 0, 0]

The middle cell of a three by three sheet sits in all four windows, so each of them holds exactly one marked cell.

Example 3

Input: m = 100000, n = 100000, coordinates = []
Output: [9999800001, 0, 0, 0, 0]

Nothing is marked, so every one of the windows holds nothing.

Constraints

2 <= m <= 10^5
2 <= n <= 10^5
0 <= coordinates.length <= 10^4
coordinates[i].length == 2
0 <= coordinates[i][0] <= m - 1
0 <= coordinates[i][1] <= n - 1
No cell is listed twice

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_black_blocks(m: int, n: int, coordinates: list[list[int]]) -> list[int]:

Java

public long[] countBlackBlocks(int m, int n, int[][] coordinates)

The intended approach

Time O(c) for c listed cells, since each touches at most four windows.Space O(c) for the tally of touched windows.

There may be ten thousand million windows, so they cannot all be visited. But a window holds a marked cell only if one of the listed cells sits in it, and each listed cell sits in at most four windows, the ones whose top-left corner is that cell or one step up, one step left, or both. So walk the listed cells and tally, for each window they touch, how many marked cells it now has, keeping the tally in a table keyed by the window's top-left corner. Only windows that lie inside the sheet count, so corners are dropped when they fall outside. The tally then gives the counts for one, two, three and four directly, and the count for none is every window less those, which is where the whole sheet's size finally comes in.

Pythonentry point: count_black_blocks

def count_black_blocks(m, n, coordinates):
    touched = {}
    for r, c in coordinates:
        for tr in (r - 1, r):
            for tc in (c - 1, c):
                if tr < 0 or tc < 0 or tr > m - 2 or tc > n - 2:
                    continue
                key = tr * n + tc
                touched[key] = touched.get(key, 0) + 1
    out = [0, 0, 0, 0, 0]
    for key in touched:
        out[touched[key]] += 1
    out[0] = (m - 1) * (n - 1) - len(touched)
    return out

The obvious version

Build the whole sheet, then visit every two-by-two window and count its marked cells one by one. Costs the size of the sheet, which is hopeless at the declared sizes but assumes nothing.

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_black_blocks(m, n, coordinates):
    marked = [[False] * n for _ in range(m)]
    for r, c in coordinates:
        marked[r][c] = True
    out = [0, 0, 0, 0, 0]
    for r in range(m - 1):
        for c in range(n - 1):
            count = 0
            if marked[r][c]:
                count += 1
            if marked[r][c + 1]:
                count += 1
            if marked[r + 1][c]:
                count += 1
            if marked[r + 1][c + 1]:
                count += 1
            out[count] += 1
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in