What is the time complexity of Minesweeper?

The optimal solution has a time complexity of O(M*N).

What data structures are used to solve Minesweeper?

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 Minesweeper?

Insight 1: Use Breadth-First Search (BFS) or Depth-First Search (DFS) to recursively reveal adjacent cells when clicking on an empty cell with no adjacent mines. Insight 2: Before revealing any cell, count the number of adjacent mines to determine the value to display or whether to trigger the recursive revealing process.

Which companies ask LeetCode 529. Minesweeper in interviews?

This problem is frequently asked at Anduril, Applied Intuition, Robinhood, Yext.

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Minesweeper - LeetCode 529 Solution

Q: What is the space complexity of Minesweeper?

The space complexity is O(M*N).

Minesweeper - Complete Solution Guide

Minesweeper is LeetCode problem 529, 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

Let's play the minesweeper game ( Wikipedia , online game )! You are given an m x n char matrix board representing the game board where: 'M' represents an unrevealed mine, 'E' represents an unrevealed empty square, 'B' represents a revealed blank square that has no adjacent mines (i.e., above, below, left, right, and all 4 diagonals), digit ( '1' to '8' ) represents how many mines are adjacent to this revealed square, and 'X' represents a revealed mine. You are also given an integer array click

Detailed Explanation

The Minesweeper problem asks us to simulate a simplified version of the Minesweeper game. We're given a 2D board represented as a character matrix and the coordinates of a click on the board. The board contains unrevealed mines ('M'), unrevealed empty squares ('E'), revealed blank squares ('B'), revealed squares with adjacent mine counts ('1' to '8'), and revealed mines ('X'). We need to update the board based on where the player clicks according to these rules: 1. If the click reveals a mine ('M'), change it to 'X' (game over). 2. If the click reveals an empty square ('E') with no adjacent mines, change it to 'B' and recursively reveal all adjacent unrevealed squares. 3. If the click reveals an empty square ('E') with at least one adjacent mine, change it to a digit representing the number of adjacent mines. 4. If neither of the above, return the board.

Solution Approach

The provided code uses Breadth-First Search (BFS) to solve the Minesweeper problem. The BFS approach starts at the clicked cell and explores its neighbors in a layer-by-layer fashion. It correctly simulates the Minesweeper game logic by checking for mines upon clicking and recursively revealing adjacent cells if an empty cell without adjacent mines is revealed. A `visited` set prevents revisiting cells, ensuring the algorithm terminates correctly.

Step-by-Step Algorithm

Step 1: Check if the clicked cell is a mine ('M'). If so, change it to 'X' and return the board.
Step 2: Initialize a queue and a visited set. Add the clicked cell to both.
Step 3: While the queue is not empty, dequeue a cell.
Step 4: For the dequeued cell, count the number of adjacent mines.
Step 5: If the mine count is greater than 0, update the cell with the mine count (converted to a char).
Step 6: If the mine count is 0, update the cell to 'B' and enqueue all unvisited adjacent empty ('E') cells, adding them to the visited set.
Step 7: Repeat steps 3-6 until the queue is empty.
Step 8: Return the updated board.

Key Insights

Insight 1: Use Breadth-First Search (BFS) or Depth-First Search (DFS) to recursively reveal adjacent cells when clicking on an empty cell with no adjacent mines.
Insight 2: Before revealing any cell, count the number of adjacent mines to determine the value to display or whether to trigger the recursive revealing process.
Insight 3: Use a `visited` set/matrix to avoid infinite loops during the recursive reveal process by ensuring each cell is visited only once.

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: Anduril, Applied Intuition, Robinhood, Yext.

Minesweeper | LeetCode 529

Medium

Array Depth-First Search Breadth-First Search Matrix

Time: O(M*N)Space: O(M*N)

Solution Code

Available in:

import java.util.LinkedList;
import java.util.Queue;
import java.util.Set;
import java.util.HashSet;

class Solution {
    public char[][] updateBoard(char[][] board, int[] click) {
        int m = board.length;
        int n = board[0].length;
        int r = click[0];
        int c = click[1];

        if (board[r][c] == 'M') {
            board[r][c] = 'X';
            return board;
        }

        Queue<int[]> q = new LinkedList<>();
        q.offer(new int[]{r, c});
        Set<String> visited = new HashSet<>();
        visited.add(r + "," + c);

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

            int mineCount = 0;
            LinkedList<int[]> neighbors = new LinkedList<>();

            for (int dr = -1; dr <= 1; dr++) {
                for (int dc = -1; dc <= 1; dc++) {
                    if (dr == 0 && dc == 0) {
                        continue;
                    }

                    int nr = row + dr;
                    int nc = col + dc;
                    if (0 <= nr && nr < m && 0 <= nc && nc < n) {
                        neighbors.add(new int[]{nr, nc});
                        if (board[nr][nc] == 'M') {
                            mineCount += 1;
                        }
                    }
                }
            }

            if (mineCount > 0) {
                board[row][col] = (char) (mineCount + '0');
            } else {
                board[row][col] = 'B';
                for (int[] neighbor : neighbors) {
                    int nr = neighbor[0];
                    int nc = neighbor[1];
                    if (board[nr][nc] == 'E' && !visited.contains(nr + "," + nc)) {
                        visited.add(nr + "," + nc);
                        q.offer(new int[]{nr, nc});
                    }
                }
            }
        }

        return board;
    }
}

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Problem Info

Description

Problem 529: Minesweeper - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Breadth-First Search Matrix

Companies

AndurilApplied IntuitionRobinhoodYext

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:529

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

8 steps

1
Step 1: Check if the clicked cell is a mine ('M'). If so, change it to 'X' and return the board.
2
Step 2: Initialize a queue and a visited set. Add the clicked cell to both.
3
Step 3: While the queue is not empty, dequeue a cell.
4
Step 4: For the dequeued cell, count the number of adjacent mines.
5
Step 5: If the mine count is greater than 0, update the cell with the mine count (converted to a char).
6
Step 6: If the mine count is 0, update the cell to 'B' and enqueue all unvisited adjacent empty ('E') cells, adding them to the visited set.
7
Step 7: Repeat steps 3-6 until the queue is empty.
8
Step 8: Return the updated board.

💡

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 O(M*N) because each cell is visited at most once. The space complexity is O(M*N) because both the queue and the visited set might hold all cells in the worst case (e.g., a board with all 'E's and the click is on one of the 'E's).

📖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 a queue, a recursive DFS approach can be used. The logic remains the same, but the order of exploration is different.

Time: O(M*N)Space: O(M*N) in the worst case due to the call stack.

Trade-offs: DFS might be slightly more concise to implement but could lead to stack overflow errors for very large boards. BFS generally provides better control over memory usage in such cases.

✨

Key Insights

1Insight 1: Use Breadth-First Search (BFS) or Depth-First Search (DFS) to recursively reveal adjacent cells when clicking on an empty cell with no adjacent mines.
2Insight 2: Before revealing any cell, count the number of adjacent mines to determine the value to display or whether to trigger the recursive revealing process.
3Insight 3: Use a `visited` set/matrix to avoid infinite loops during the recursive reveal process by ensuring each cell is visited only once.

📚

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: Clicking on a mine ('M'). The solution correctly handles this by changing the cell to 'X' and returning the board immediately.
Edge case 2: Clicking on an 'E' surrounded by mines. The solution correctly calculates the adjacent mine count and updates the cell with the appropriate digit.
Edge case 3: Clicking near the boundaries of the board. The solution correctly checks the boundaries before accessing adjacent cells, avoiding out-of-bounds errors.
Edge case 4: Clicking on a cell that has already been revealed. The algorithm skips processing it since the `visited` set will contain the cell and the condition `board[nr][nc] == 'E' && !visited.contains(nr +
+ nc)` will not allow to enqueue the cell again

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking for out-of-bounds errors when accessing adjacent cells, leading to runtime errors.

✗Common mistake 2: Failing to use a `visited` set or matrix, leading to infinite loops when recursively revealing adjacent cells.

✗Common mistake 3: Incorrectly calculating the adjacent mine count.

✗Common mistake 4: Not handling the case when the first click reveals a mine.

✅

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: This algorithm can be used in actual Minesweeper games or similar grid-based puzzle games.

🏦Real-world scenario 2: Flood fill algorithms used in image editing software share similar logic of exploring and modifying connected regions. This algorithm helps understanding the concepts used in flood fill.

🎮Real-world scenario 3: Pathfinding algorithms in games also employ similar exploration techniques to find optimal routes.

💡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.

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: November 17, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minesweeper"?

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.

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