How do you solve Number of Black Blocks on LeetCode?

Number of Black Blocks (LeetCode #2768) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashSet allows for quick lookups of black cells.

What is the brute force solution for Number of Black Blocks?

The brute force approach for Number of Black Blocks is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible 2x2 block in the grid to count how many black cells it contains. This approach is straightforward but inefficient for larger grids. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Black Blocks?

Number of Black Blocks uses the following concepts: Array, Hash Table, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Black Blocks?

Always clarify the problem constraints and edge cases. Think about space and time efficiency, especially with large inputs. Explain your thought process clearly while coding.

What are common mistakes when solving Number of Black Blocks?

Not considering edge cases where blocks might go out of bounds. Overlooking the need to count blocks with 0 black cells.

#2768

Number of Black Blocks

Medium

ArrayHash TableEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking every block, we can use a HashSet to track the black cells and count the blocks based on their contributions to neighboring blocks. This reduces unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Initialize a HashSet to store the black cell coordinates.
2Step 2: For each black cell, determine which 2x2 blocks it contributes to.
3Step 3: For each block, count the number of black cells it contains based on the black cells that contribute to it.
4Step 4: Update the count array for blocks with 0 to 4 black cells.

solution.py15 lines

1# Full working Python code
2
3def countBlackBlocks(m, n, coordinates):
4    black_cells = set(map(tuple, coordinates))
5    count = [0] * 5
6    for x, y in black_cells:
7        for dx in range(2):
8            for dy in range(2):
9                block_x, block_y = x - dx, y - dy
10                if 0 <= block_x < m - 1 and 0 <= block_y < n - 1:
11                    black_count = sum((block_x + i, block_y + j) in black_cells for i in range(2) for j in range(2))
12                    count[black_count] += 1
13    count[0] = (m - 1) * (n - 1) - sum(count)  # Calculate blocks with 0 black cells
14    return count
15

ℹ

Complexity note: The time complexity is O(n) because we only iterate over the black cells and their contributions to blocks. The space complexity is O(n) due to storing the black cells in a HashSet.

1Using a HashSet allows for quick lookups of black cells.
2Counting contributions from black cells reduces unnecessary checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.