2768. Number of Black Blocks

Problem Link

Description

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You are given two integers m and n representing the dimensions of a 0-indexed m x n grid.

You are also given a 0-indexed 2D integer matrix coordinates, where coordinates[i] = [x, y] indicates that the cell with coordinates [x, y] is colored black. All cells in the grid that do not appear in coordinates are white.

A block is defined as a 2 x 2 submatrix of the grid. More formally, a block with cell [x, y] as its top-left corner where 0 <= x < m - 1 and 0 <= y < n - 1 contains the coordinates [x, y], [x + 1, y], [x, y + 1], and [x + 1, y + 1].

Return a 0-indexed integer array arr of size 5 such that arr[i] is the number of blocks that contains exactly i black cells.

 

Example 1:

Input: m = 3, n = 3, coordinates = [[0,0]]
Output: [3,1,0,0,0]
Explanation: The grid looks like this:

There is only 1 block with one black cell, and it is the block starting with cell [0,0].
The other 3 blocks start with cells [0,1], [1,0] and [1,1]. They all have zero black cells. 
Thus, we return [3,1,0,0,0]. 

Example 2:

Input: m = 3, n = 3, coordinates = [[0,0],[1,1],[0,2]]
Output: [0,2,2,0,0]
Explanation: The grid looks like this:

There are 2 blocks with two black cells (the ones starting with cell coordinates [0,0] and [0,1]).
The other 2 blocks have starting cell coordinates of [1,0] and [1,1]. They both have 1 black cell.
Therefore, we return [0,2,2,0,0].

 

Constraints:

• 2 <= m <= 105
• 2 <= n <= 105
• 0 <= coordinates.length <= 104
• coordinates[i].length == 2
• 0 <= coordinates[i][0] < m
• 0 <= coordinates[i][1] < n
• It is guaranteed that coordinates contains pairwise distinct coordinates.

Solution

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Python3

class Solution:
    def countBlackBlocks(self, rows: int, cols: int, cord: List[List[int]]) -> List[int]:
        res = [0] * 5
        s = set(map(tuple, cord))
        seen = set()
        
        for x, y in cord:
            for dx, dy in [[x, y], [x - 1, y], [x, y - 1], [x - 1, y - 1]]:
                if 0 <= dx < rows - 1 and 0 <= dy < cols - 1 and (dx, dy) not in seen:
                    curr = 0
                    
                    for nx, ny in [[dx, dy], [dx + 1, dy], [dx, dy + 1], [dx + 1, dy + 1]]:
                        if (nx, ny) in s:
                            curr += 1
                    
                    if curr > 0:
                        res[curr] += 1
                        
                    seen.add((dx, dy))
        
        res[0] = (rows - 1) * (cols - 1) - sum(res[1:])
        
        return res