What is the time complexity of Surrounded Regions?

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

What is the space complexity of Surrounded Regions?

The space complexity is O(m*n).

What data structures are used to solve Surrounded Regions?

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

What are the key insights for solving Surrounded Regions?

Insight 1: The key is to identify 'O's that are *not* surrounded. These are the 'O's that are either on the border or connected to 'O's on the border. Insight 2: A temporary marker is needed to differentiate between 'O's that are connected to the border ('safe' 'O's) and 'O's that are truly surrounded.

Which companies ask LeetCode 130. Surrounded Regions in interviews?

This problem is frequently asked at Arcesium, Flipkart, Nutanix, TikTok, Urban Company.

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Surrounded Regions - LeetCode 130 Solution

Surrounded Regions - Complete Solution Guide

Surrounded Regions is LeetCode problem 130, 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 matrix board containing letters 'X' and 'O' , capture regions that are surrounded : Connect : A cell is connected to adjacent cells horizontally or vertically. Region : To form a region connect every 'O' cell. Surround : The region is surrounded with 'X' cells if you can connect the region with 'X' cells and none of the region cells are on the edge of the board . To capture a surrounded region , replace all 'O' s with 'X' s in-place within the original board. You do not ne

Detailed Explanation

The problem asks us to capture regions of 'O's in a 2D board that are completely surrounded by 'X's. Capturing a region means replacing all 'O's within that region with 'X's. A region of 'O's is considered surrounded if it's fully enclosed by 'X's and none of the 'O's in that region are connected to the edge of the board. The input is a 2D character array (board), and the output is the modified board (in-place).

Solution Approach

The solution utilizes Depth-First Search (DFS) to identify and mark 'O's that are connected to the border. It then iterates through the board, flipping the remaining 'O's to 'X's (the surrounded regions) and reverting the marked 'O's back to 'O's.

Step-by-Step Algorithm

Step 1: Iterate through the borders of the board (top, bottom, left, right).
Step 2: For each 'O' found on the border, initiate a DFS traversal.
Step 3: The DFS function recursively explores adjacent cells. If an adjacent cell is within bounds and is an 'O', it is marked with a temporary character (e.g., 'T'). The DFS continues recursively from this marked cell.
Step 4: After the border traversal and DFS marking, iterate through the entire board.
Step 5: If a cell contains 'O', it means it was not connected to any border 'O' and thus is surrounded. Change it to 'X'.
Step 6: If a cell contains the temporary character (e.g., 'T'), it means it was connected to a border 'O' and is not surrounded. Change it back to 'O'.

Key Insights

Insight 1: The key is to identify 'O's that are *not* surrounded. These are the 'O's that are either on the border or connected to 'O's on the border.
Insight 2: A temporary marker is needed to differentiate between 'O's that are connected to the border ('safe' 'O's) and 'O's that are truly surrounded.
Insight 3: Depth-First Search (DFS) is an efficient way to traverse and mark all 'O's connected to the border 'O's.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

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

Companies

Asked at: Arcesium, Flipkart, Nutanix, TikTok, Urban Company.

Surrounded Regions | LeetCode 130

Medium

Array Depth-First Search Breadth-First Search Union Find Matrix

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

Solution Code

Available in:

class Solution {
    public void solve(char[][] board) {
        if (board == null || board.length == 0) {
            return;
        }

        int rows = board.length;
        int cols = board[0].length;

        // Mark 'O's connected to the top and bottom borders
        for (int j = 0; j < cols; j++) {
            if (board[0][j] == 'O') {
                dfs(board, 0, j);
            }
            if (board[rows - 1][j] == 'O') {
                dfs(board, rows - 1, j);
            }
        }

        // Mark 'O's connected to the left and right borders
        for (int i = 0; i < rows; i++) {
            if (board[i][0] == 'O') {
                dfs(board, i, 0);
            }
            if (board[i][cols - 1] == 'O') {
                dfs(board, i, cols - 1);
            }
        }

        // Flip 'O's to 'X's and 'T's back to 'O's
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                } else if (board[i][j] == 'T') {
                    board[i][j] = 'O';
                }
            }
        }
    }

    private void dfs(char[][] board, int row, int col) {
        int rows = board.length;
        int cols = board[0].length;

        if (row < 0 || row >= rows || col < 0 || col >= cols || board[row][col] != 'O') {
            return;
        }

        board[row][col] = 'T'; // Mark as temporary

        dfs(board, row + 1, col);
        dfs(board, row - 1, col);
        dfs(board, row, col + 1);
        dfs(board, row, col - 1);
    }
}

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Problem Info

Description

Problem 130: Surrounded Regions - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Breadth-First Search Union Find Matrix

Companies

ArcesiumFlipkartNutanixTikTokUrban Company

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:130

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

6 steps

1
Step 1: Iterate through the borders of the board (top, bottom, left, right).
2
Step 2: For each 'O' found on the border, initiate a DFS traversal.
3
Step 3: The DFS function recursively explores adjacent cells. If an adjacent cell is within bounds and is an 'O', it is marked with a temporary character (e.g., 'T'). The DFS continues recursively from this marked cell.
4
Step 4: After the border traversal and DFS marking, iterate through the entire board.
5
Step 5: If a cell contains 'O', it means it was not connected to any border 'O' and thus is surrounded. Change it to 'X'.
6
Step 6: If a cell contains the temporary character (e.g., 'T'), it means it was connected to a border 'O' and is not surrounded. Change it back to 'O'.

💡

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 time complexity is derived from the nested loops used to traverse the board and the DFS, which in the worst-case scenario can visit every cell. The space complexity comes primarily from the DFS recursion stack, which can be proportional to the number of cells on the board in the worst case. While the code doesn't use any extra data structures that scale with the input size beyond the modified board, the implicit stack space of the recursive DFS can become significant.

📖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

Breadth-First Search (BFS)

Instead of DFS, BFS can be used to traverse and mark connected 'O's. Starting from the border 'O's, BFS uses a queue to explore neighboring 'O's level by level.

Time: O(m*n)Space: O(m*n) in the worst case (when almost all cells are 'O's).

Trade-offs: BFS might be slightly more efficient than DFS in some cases due to iterative nature but uses a queue which takes up space. DFS uses recursion, potentially limited by the call stack size. For this problem, both are acceptable, with DFS being slightly more concise.

Union Find (Disjoint Set)

Use a Union Find data structure to connect all 'O's to their neighbors. Then, create a dummy 'O' node representing the border. Connect all border 'O's to the dummy 'O' node. Finally, iterate through the board. If an 'O' is not connected to the dummy 'O' node, it is surrounded and should be flipped.

Time: O(m*n * α(m*n)), where α is the inverse Ackermann function, which grows very slowly, so it's essentially O(m*n).Space: O(m*n) to store the parent array for the Union Find data structure.

Trade-offs: Union Find is generally more complex to implement than DFS or BFS for this specific problem. While it has similar time complexity asymptotically, the constant factors and implementation overhead might make it less efficient in practice. It can be more advantageous if there are more complex connectivity scenarios or queries beyond just flipping surrounded regions.

✨

Key Insights

1Insight 1: The key is to identify 'O's that are *not* surrounded. These are the 'O's that are either on the border or connected to 'O's on the border.
2Insight 2: A temporary marker is needed to differentiate between 'O's that are connected to the border ('safe' 'O's) and 'O's that are truly surrounded.
3Insight 3: Depth-First Search (DFS) is an efficient way to traverse and mark all 'O's connected to the border 'O's.

📚

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: Empty board. The solution handles this by checking if the board is null or empty at the beginning and returning if so, preventing a null pointer exception.
Edge case 2: Board with a single 'X'. The code correctly handles this as no 'O' exists, and nothing will be changed.
Edge case 3: Board with a single 'O'. If the 'O' is on the border, it will be marked and then restored. If the 'O' is not on the border, it will be changed to an 'X'.
Edge case 4: Board with all 'X's. No 'O's will be found, so nothing will be modified, which is correct.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge cases where the board is empty or has only one row/column. These cases need to be explicitly checked to avoid errors.

✗Common mistake 2: Forgetting to restore the temporary marker (e.g., 'T') back to 'O' after flipping the surrounded 'O's to 'X's. This results in an incorrect final board state.

✗Common mistake 3: Not considering the boundary conditions correctly in DFS/BFS, which may lead to index-out-of-bounds errors.

✗Common mistake 4: Incorrectly identifying the surrounded regions and flipping 'O's that are actually connected to the border.

✅

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 and filling enclosed regions in images. This could be useful for object detection or segmentation.

🏦Real-world scenario 2: Game development - Determining if a section of a game map is enclosed by walls (e.g., in a strategy game) or if a player is 'trapped'.

🎮Real-world scenario 3: Network connectivity analysis - Determining if a node in a network is isolated from the external network or is part of a larger connected component.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 9, 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

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

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 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 "Surrounded Regions"?

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?