What is the time complexity of Coloring A Border?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Coloring A Border?

The space complexity is O(m*n).

What data structures are used to solve Coloring A Border?

This problem uses Array, Depth-First Search, Breadth-First Search, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Coloring A Border?

Insight 1: We need to identify the connected component using either Depth-First Search (DFS) or Breadth-First Search (BFS). Insight 2: During the search, we need to determine whether a cell is part of the border, meaning it's either at the grid boundary or adjacent to a cell with a different color.

Which companies ask LeetCode 1034. Coloring A Border in interviews?

This problem is frequently asked at Booking.com.

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Coloring A Border - LeetCode 1034 Solution

Coloring A Border - Complete Solution Guide

Coloring A Border is LeetCode problem 1034, 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 an m x n integer matrix grid , and three integers row , col , and color . Each value in the grid represents the color of the grid square at that location. Two squares are called adjacent if they are next to each other in any of the 4 directions. Two squares belong to the same connected component if they have the same color and they are adjacent. The border of a connected component is all the squares in the connected component that are either adjacent to (at least) a square not in t

Detailed Explanation

The problem asks us to color the border of a connected component in a given grid. A connected component is a group of adjacent squares with the same color. The border of a connected component consists of squares that are either adjacent to a square outside the component (different color or out of bounds) or located on the edge of the grid. Given a starting cell (row, col) and a new color, we must change the color of the border of the connected component containing that starting cell to the new color. The input is the grid (a 2D array of integers representing colors), the starting row and column indices, and the new color. The output is the modified grid with the border colored.

Solution Approach

The provided code uses Breadth-First Search (BFS) to traverse the connected component starting from the given row and column. During the traversal, it checks for each cell if it is a border cell. If it is, the cell is added to a list of border squares. After the BFS is complete, it iterates through the list of border squares and changes their colors to the new color. This ensures that we change the color of all border squares in the connected component.

Step-by-Step Algorithm

Step 1: Initialize a queue with the starting cell (row, col).
Step 2: Initialize a visited set to keep track of visited cells to avoid cycles.
Step 3: Initialize an empty list to store the border squares.
Step 4: While the queue is not empty, dequeue a cell (r, c).
Step 5: Check if the cell is a border cell by examining its four neighbors (up, down, left, right). A cell is a border cell if any of its neighbors are out of bounds or have a different color than the initial color.
Step 6: If the cell is a border cell, add it to the list of border squares.
Step 7: For each valid neighbor with the same initial color that hasn't been visited, enqueue it and add it to the visited set.
Step 8: After the BFS is complete, iterate through the list of border squares and change their colors to the new color.
Step 9: Return the modified grid.

Key Insights

Insight 1: We need to identify the connected component using either Depth-First Search (DFS) or Breadth-First Search (BFS).
Insight 2: During the search, we need to determine whether a cell is part of the border, meaning it's either at the grid boundary or adjacent to a cell with a different color.
Insight 3: We can use a visited set to keep track of the cells we have already explored in the connected component to prevent infinite loops.
Insight 4: Handling the case where the initial color is the same as the new color is important for efficiency. The solution can simply return the original grid in this scenario.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Depth-First Search, Breadth-First Search, Matrix.

Companies

Asked at: Booking.com.

Coloring A Border | LeetCode 1034

Medium

Array Depth-First Search Breadth-First Search Matrix

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

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Queue;
import java.util.Set;

class Solution {
    public int[][] colorBorder(int[][] grid, int row, int col, int color) {
        int m = grid.length;
        int n = grid[0].length;
        int initial_color = grid[row][col];

        if (initial_color == color) {
            return grid;
        }

        Queue<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {row, col});
        Set<String> visited = new HashSet<>();
        visited.add(row + "," + col);
        List<int[]> border_squares = new ArrayList<>();

        while (!q.isEmpty()) {
            int[] curr = q.poll();
            int r = curr[0];
            int c = curr[1];

            boolean is_border = false;
            int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
            for (int[] dir : directions) {
                int dr = dir[0];
                int dc = dir[1];
                int nr = r + dr;
                int nc = c + dc;

                if (0 <= nr && nr < m && 0 <= nc && nc < n && grid[nr][nc] == initial_color) {
                    if (!visited.contains(nr + "," + nc)) {
                        visited.add(nr + "," + nc);
                        q.offer(new int[] {nr, nc});
                    }
                } else {
                    is_border = true;
                }
            }

            if (is_border) {
                border_squares.add(new int[] {r, c});
            }
        }

        for (int[] square : border_squares) {
            int r = square[0];
            int c = square[1];
            grid[r][c] = color;
        }

        return grid;
    }
}

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Problem Info

Description

Problem 1034: Coloring A Border - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Breadth-First Search Matrix

Companies

Booking.com

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1034

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 queue with the starting cell (row, col).
2
Step 2: Initialize a visited set to keep track of visited cells to avoid cycles.
3
Step 3: Initialize an empty list to store the border squares.
4
Step 4: While the queue is not empty, dequeue a cell (r, c).
5
Step 5: Check if the cell is a border cell by examining its four neighbors (up, down, left, right). A cell is a border cell if any of its neighbors are out of bounds or have a different color than the initial color.
6
Step 6: If the cell is a border cell, add it to the list of border squares.
7
Step 7: For each valid neighbor with the same initial color that hasn't been visited, enqueue it and add it to the visited set.
8
Step 8: After the BFS is complete, iterate through the list of border squares and change their colors to the new color.
9
Step 9: Return the modified grid.

💡

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(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

The BFS algorithm visits each cell at most once, which takes O(1) time per cell. The space is mainly determined by the queue and the visited set, which can store up to m*n elements in the worst-case scenario (entire grid is the component).

📖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

Depth-First Search (DFS)

Instead of using BFS, we could use DFS to traverse the connected component. The logic for identifying border cells remains the same.

Time: O(m*n)Space: O(m*n) in the worst case (for the recursion stack).

Trade-offs: DFS may use more stack space than BFS in some cases, especially for large connected components. However, DFS can sometimes be easier to implement recursively.

✨

Key Insights

1Insight 1: We need to identify the connected component using either Depth-First Search (DFS) or Breadth-First Search (BFS).
2Insight 2: During the search, we need to determine whether a cell is part of the border, meaning it's either at the grid boundary or adjacent to a cell with a different color.
3Insight 3: We can use a visited set to keep track of the cells we have already explored in the connected component to prevent infinite loops.
4Insight 4: Handling the case where the initial color is the same as the new color is important for efficiency. The solution can simply return the original grid in this scenario.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: The grid is empty (m=0 or n=0). Although the constraints specify that m and n are at least 1, it's good practice to consider this edge case for robustness. The provided solutions implicitly handle this as the loops won't execute.
Edge case 2: The starting cell is already the desired color. In this case, no changes are needed, so the original grid is returned directly.
Edge case 3: The connected component occupies the entire grid. In this scenario, all boundary cells need to be colored. The algorithm correctly identifies all border cells in this situation.
Edge case 4: The starting cell is on the boundary of the grid. This case is correctly handled by the boundary checks within the BFS algorithm.
Edge case 5: The grid contains only one cell. The algorithm correctly identifies if this cell is a border and colors it if needed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking the bounds of the grid when accessing neighbors. This can lead to array out-of-bounds exceptions.

✗Common mistake 2: Not keeping track of visited cells, leading to infinite loops.

✗Common mistake 3: Incorrectly identifying border cells. For example, forgetting to check if a neighbor is out of bounds or has a different color.

✗Common mistake 4: Modifying the grid while traversing it, which can lead to incorrect results. It's better to store the border cells and then update the grid after the traversal is complete.

✅

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, where you might want to identify and highlight the boundaries of objects in an image.

🏦Real-world scenario 2: Flood fill algorithms, used in paint programs to fill connected regions with a specific color. The border detection part is crucial for restricting the filled area.

🎮Real-world scenario 3: Network analysis, where you might want to identify the boundary nodes of a connected network component.

📱Real-world scenario 4: Maze solving, where finding the 'border' could be useful in finding exits and dead ends.

🚗Real-world scenario 5: Map coloring - identifying adjacent regions with the same property to re-color them based on some criteria.

💡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: October 13, 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

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

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

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 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 "Coloring A Border"?

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 depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?