What is the time complexity of Number of Black Blocks?

The optimal solution has a time complexity of O(K).

What data structures are used to solve Number of Black Blocks?

This problem uses Array, Hash Table, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Black Blocks?

Insight 1: Instead of iterating through all possible 2x2 blocks, we only need to focus on the blocks adjacent to the given black cells. This significantly reduces the number of blocks to check. Insight 2: Using a hash table (or dictionary) to store and update the counts of black cells in each affected block is crucial for efficient counting.

Which companies ask LeetCode 2768. Number of Black Blocks in interviews?

This problem is frequently asked at Block, Capital One, SIG, Visa, X.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Number of Black Blocks - LeetCode 2768 Solution

Q: How to solve Number of Black Blocks?

The solution iterates through the given black cell coordinates. For each black cell at (r, c), it checks the 2x2 blocks that could potentially contain this cell. These are the blocks with top-left corners at (r-1, c-1), (r-1, c), (r, c-1), and (r, c). For each of these potential blocks, it verifies if the block is within the grid boundaries. If it is, it increments the black cell count for that block using a hash table. After processing all the black cells, it calculates the total number of blocks, counts the number of affected blocks (those in the hash table), and calculates the number of blocks with zero black cells. Finally, it iterates through the hash table, counts the number of blocks with 1 to 4 black cells, and returns the result.

Number of Black Blocks - Complete Solution Guide

Number of Black Blocks is LeetCode problem 2768, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 [

Detailed Explanation

The problem asks us to analyze a grid of size m x n where some cells are colored black based on given coordinates. We need to count the number of 2x2 blocks (submatrices) that contain exactly 0, 1, 2, 3, or 4 black cells. The input is the grid dimensions (m, n) and a list of coordinates representing the black cells. The output is an array of size 5, where each element represents the count of blocks with the corresponding number of black cells.

Solution Approach

The solution iterates through the given black cell coordinates. For each black cell at (r, c), it checks the 2x2 blocks that could potentially contain this cell. These are the blocks with top-left corners at (r-1, c-1), (r-1, c), (r, c-1), and (r, c). For each of these potential blocks, it verifies if the block is within the grid boundaries. If it is, it increments the black cell count for that block using a hash table. After processing all the black cells, it calculates the total number of blocks, counts the number of affected blocks (those in the hash table), and calculates the number of blocks with zero black cells. Finally, it iterates through the hash table, counts the number of blocks with 1 to 4 black cells, and returns the result.

Step-by-Step Algorithm

Step 1: Initialize a hash table (or dictionary) to store the count of black cells in each 2x2 block.
Step 2: Iterate through the given coordinates of black cells.
Step 3: For each black cell (r, c), iterate through the possible top-left corners of 2x2 blocks that may include it: (r-1, c-1), (r-1, c), (r, c-1), and (r, c).
Step 4: For each potential top-left corner (i, j), check if it is a valid block within the grid boundaries (0 <= i < m - 1 and 0 <= j < n - 1).
Step 5: If the block is valid, increment the count of black cells for that block in the hash table.
Step 6: After processing all black cells, calculate the total number of possible 2x2 blocks: (m - 1) * (n - 1).
Step 7: Calculate the number of blocks with 0 black cells: total blocks - number of affected blocks (size of the hash table).
Step 8: Iterate through the hash table and count the blocks with 1, 2, 3, and 4 black cells.
Step 9: Return the array containing the counts of blocks with 0, 1, 2, 3, and 4 black cells.

Key Insights

Insight 1: Instead of iterating through all possible 2x2 blocks, we only need to focus on the blocks adjacent to the given black cells. This significantly reduces the number of blocks to check.
Insight 2: Using a hash table (or dictionary) to store and update the counts of black cells in each affected block is crucial for efficient counting.
Insight 3: After counting the blocks with 1 to 4 black cells, the count of blocks with 0 black cells can be efficiently calculated by subtracting the sum of counts of blocks with 1 to 4 black cells from the total number of possible blocks (m-1)*(n-1).

Complexity Analysis

Time Complexity: O(K)

Space Complexity: O(K)

Topics

This problem involves: Array, Hash Table, Enumeration.

Companies

Asked at: Block, Capital One, SIG, Visa, X.

Number of Black Blocks | LeetCode 2768

Medium

Array Hash Table Enumeration

Time: O(K)Space: O(K)

Solution Code

Available in:

import java.util.*;

class Solution {
    public long[] countBlackBlocks(int m, int n, int[][] coordinates) {
        Map<Pair, Integer> block_counts = new HashMap<>();

        for (int[] coordinate : coordinates) {
            int r = coordinate[0];
            int c = coordinate[1];
            for (int i = r - 1; i <= r; i++) {
                for (int j = c - 1; j <= c; j++) {
                    if (0 <= i && i < m - 1 && 0 <= j && j < n - 1) {
                        Pair p = new Pair(i, j);
                        block_counts.put(p, block_counts.getOrDefault(p, 0) + 1);
                    }
                }
            }
        }

        long[] ans = new long[5];

        long total_blocks = (long) (m - 1) * (n - 1);

        long num_affected_blocks = block_counts.size();

        ans[0] = total_blocks - num_affected_blocks;

        for (int count : block_counts.values()) {
            if (count <= 4) {
                ans[count] += 1;
            }
        }

        return ans;
    }

    private static class Pair {
        int x, y;

        public Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return x == pair.x && y == pair.y;
        }

        @Override
        public int hashCode() {
            return Objects.hash(x, y);
        }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2768: Number of Black Blocks - Detailed explanation and solution approaches.

Topics

Array Hash Table Enumeration

Companies

BlockCapital OneSIGVisaX

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:2768

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize a hash table (or dictionary) to store the count of black cells in each 2x2 block.
2
Step 2: Iterate through the given coordinates of black cells.
3
Step 3: For each black cell (r, c), iterate through the possible top-left corners of 2x2 blocks that may include it: (r-1, c-1), (r-1, c), (r, c-1), and (r, c).
4
Step 4: For each potential top-left corner (i, j), check if it is a valid block within the grid boundaries (0 <= i < m - 1 and 0 <= j < n - 1).
5
Step 5: If the block is valid, increment the count of black cells for that block in the hash table.
6
Step 6: After processing all black cells, calculate the total number of possible 2x2 blocks: (m - 1) * (n - 1).
7
Step 7: Calculate the number of blocks with 0 black cells: total blocks - number of affected blocks (size of the hash table).
8
Step 8: Iterate through the hash table and count the blocks with 1, 2, 3, and 4 black cells.
9
Step 9: Return the array containing the counts of blocks with 0, 1, 2, 3, and 4 black cells.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(K)

💾

Space Complexity

O(K)

📝

Why?

The dominant factor in both time and space complexity is the number of coordinates (K). The algorithm processes each coordinate a constant number of times. The hash table's size is bounded by a multiple of K because each coordinate can affect at most four blocks.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible 2x2 blocks in the grid and count the number of black cells in each block.

Time: O(m * n)Space: O(1)

Trade-offs: This approach has a higher time complexity than the main solution and is less efficient for large grids and a relatively small number of black cells. However, it uses constant space.

✨

Key Insights

1Insight 1: Instead of iterating through all possible 2x2 blocks, we only need to focus on the blocks adjacent to the given black cells. This significantly reduces the number of blocks to check.
2Insight 2: Using a hash table (or dictionary) to store and update the counts of black cells in each affected block is crucial for efficient counting.
3Insight 3: After counting the blocks with 1 to 4 black cells, the count of blocks with 0 black cells can be efficiently calculated by subtracting the sum of counts of blocks with 1 to 4 black cells from the total number of possible blocks (m-1)*(n-1).

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Empty coordinates array. The code correctly handles this by initializing the counts for all blocks to 0, and then calculating the number of blocks with 0 black cells as (m-1)*(n-1).
Edge case 2: Coordinates at the boundary of the grid. The code correctly checks for valid block indices (0 <= i < m - 1 and 0 <= j < n - 1) and avoids accessing out-of-bounds cells.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the boundaries of the 2x2 blocks. Off-by-one errors can occur when determining the top-left corner coordinates of the blocks adjacent to the black cells.

✗Common mistake 2: Forgetting to check for out-of-bounds block indices. This can lead to errors when accessing grid cells outside the valid range.

✗Common mistake 3: Double-counting blocks. Using a hash table prevents this, but if a different data structure is used without proper checks, blocks could be counted multiple times.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Image processing. Identifying patterns or features in a large image by analyzing smaller subregions (similar to the 2x2 blocks). For example, identifying edges or corners by counting the number of 'on' pixels in a small area.

🏦Real-world scenario 2: Game development. Analyzing game boards to identify clusters of similar elements or patterns. For example, in a grid-based game, counting the number of adjacent enemy units to determine threat levels.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Black Blocks"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?