LeetCode 529: Minesweeper Solution in Python – A Step-by-Step Guide

What Is LeetCode 529: Minesweeper?

In LeetCode 529: Minesweeper, you’re given a 2D character board representing a Minesweeper game and a click position [click[0], click[1]]. The board contains mines ('M'), empty cells ('E'), and revealed cells ('B' for blank, digits for mine counts). Your task is to update the board after the click based on Minesweeper rules: reveal mines, count adjacent mines, or recursively reveal empty cells. For example, clicking an empty cell might reveal a number or open more cells. This problem builds on LeetCode 200: Number of Islands for grid traversal.

Problem Statement

• Input:
• board (List[List[str]]): Minesweeper grid.
• click (List[int]): [row, col] of click position.
• Output: Updated board after click.
• Rules:
• If mine clicked: 'M' → 'X'.
• If empty clicked: Reveal adjacent mine count or recursively reveal if no mines nearby.
• Adjacent = 8 surrounding cells.

Constraints

• 1 <= board.length, board[i].length <= 50
• board[i][j] is 'M', 'E', 'B', or digit 1-8.
• 0 <= click[0], click[1] < board.length
• Click position is valid.

Examples

• Input: board = [["E","E","E","E","E"],["E","E","M","E","E"],["E","E","E","E","E"],["E","E","E","E","E"]], click = [3,0]
• Output: [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","E","1","B"],["B","B","E","B","B"]]
• Clicked empty cell reveals recursively.
• Input: board = [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","E","1","B"],["B","B","E","B","B"]], click = [1,2]
• Output: [["B","1","E","1","B"],["B","1","X","1","B"],["B","1","E","1","B"],["B","B","E","B","B"]]
• Clicked mine turns to 'X'.

Understanding the Problem: Sweeping the Board

To solve LeetCode 529: Minesweeper in Python, we need to update a Minesweeper board after a click, following game rules: reveal a mine as 'X', or for an empty cell, show the count of adjacent mines and possibly reveal more cells. A naive approach might check each cell individually, but with recursive or iterative traversal, we can optimize. We’ll explore:

• Best Solution (DFS with Recursive Reveal): O(m n) time, O(m n) space—fast and natural.
• Alternative Solution (BFS with Queue): O(m n) time, O(m n) space—systematic but bulkier.

Let’s dive into the DFS solution with a friendly breakdown!

Best Solution: DFS with Recursive Reveal

Why DFS Wins

The DFS with recursive reveal is the best for LeetCode 529 because it mirrors Minesweeper’s natural behavior—clicking an empty cell recursively reveals connected empty areas—running in O(m * n) time and O(m * n) space due to recursion stack, with a clean, intuitive implementation. It’s like clicking a cell and watching the board unfold!

How It Works

Think of this as a Minesweeper ripple effect:

• Step 1: Handle Mine Click:
• If clicked cell is 'M', set to 'X', return board.
• Step 2: DFS for Empty Cell:
• If 'E', count adjacent mines.
• If 0 mines, set to 'B', recursively reveal 8 neighbors.
• If >0 mines, set to count, stop.
• Step 3: Base Case:
• Skip non-'E' cells or out-of-bounds.
• Why It Works:
• DFS mimics game’s recursive reveal.
• Counts mines efficiently per cell.

It’s like a Minesweeper reveal wizard!

Step-by-Step Example

Example: board = [["E","E","E"],["E","M","E"],["E","E","E"]], click = [0,0]

• Init: Click [0,0] = 'E'.
• Step 1: Not mine, proceed.
• Step 2: DFS at (0,0):
• Count mines: Check 8 neighbors (3 in bounds: (0,1), (1,0), (1,1)=M), 1 mine.
• Set board[0][0] = '1', stop (no recursion).
• Result: [["1","E","E"],["E","M","E"],["E","E","E"]].

Example: click = [2,0]

• DFS at (2,0):
• 0 mines (neighbors: (1,0), (1,1)=M, (2,1)), set 'B'.
• Recurse:
• (1,0): 1 mine (1,1=M), '1'.
• (2,1): 1 mine (1,1=M), '1'.
• Result: [["E","E","E"],["1","M","E"],["B","1","E"]].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        # Step 1: Handle mine click
        row, col = click
        if board[row][col] == 'M':
            board[row][col] = 'X'
            return board

        # Step 2: Directions for 8 neighbors
        directions = [(-1,-1), (-1,0), (-1,1), (0,-1), (0,1), (1,-1), (1,0), (1,1)]

        def dfs(r, c):
            # Base case: Out of bounds or not 'E'
            if (r < 0 or r >= len(board) or c < 0 or c >= len(board[0]) or 
                board[r][c] != 'E'):
                return

            # Count adjacent mines
            mine_count = 0
            for dr, dc in directions:
                nr, nc = r + dr, c + dc
                if (0 <= nr < len(board) and 0 <= nc < len(board[0]) and 
                    board[nr][nc] == 'M'):
                    mine_count += 1

            # Update cell and recurse if no mines
            if mine_count == 0:
                board[r][c] = 'B'
                for dr, dc in directions:
                    dfs(r + dr, c + dc)
            else:
                board[r][c] = str(mine_count)

        # Step 3: Start DFS from click
        dfs(row, col)
        return board

• Lines 4-7: If mine clicked, mark 'X', return.
• Line 10: 8 directions for neighbors.
• Lines 13-15: DFS base case: skip invalid or non-empty cells.
• Lines 18-23: Count mines in 8 neighbors.
• Lines 26-30: If 0 mines, set 'B', recurse; else set count.
• Line 33: Start DFS at click.
• Time Complexity: O(m * n)—visit each cell once.
• Space Complexity: O(m * n)—recursion stack.

It’s like a Minesweeper ripple master!

Alternative Solution: BFS with Queue

Why an Alternative Approach?

The BFS with queue solution processes cells level-by-level using a queue, ensuring all neighbors are revealed systematically. It’s O(m * n) time and O(m * n) space—straightforward but uses extra queue space. Great for iterative fans!

How It Works

Picture this as a Minesweeper wave:

• Step 1: Handle mine click as above.
• Step 2: BFS for empty cell:
• Queue starts with clicked cell.
• Process each cell, count mines, reveal or stop.
• Step 3: Use queue to manage neighbors.

It’s like a Minesweeper wave spreader!

Step-by-Step Example

Example: Same board, click = [2,0]

• Init: Queue = [(2,0)].
• BFS:
• (2,0): 0 mines, 'B', enqueue (1,0), (2,1).
• (1,0): 1 mine, '1'.
• (2,1): 1 mine, '1'.
• Result: Same as DFS.

Code for BFS Approach

Copy codefrom collections import deque

class Solution:
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        row, col = click
        if board[row][col] == 'M':
            board[row][col] = 'X'
            return board

        directions = [(-1,-1), (-1,0), (-1,1), (0,-1), (0,1), (1,-1), (1,0), (1,1)]
        queue = deque([(row, col)])

        while queue:
            r, c = queue.popleft()
            if board[r][c] != 'E':
                continue

            mine_count = 0
            for dr, dc in directions:
                nr, nc = r + dr, c + dc
                if 0 <= nr < len(board) and 0 <= nc < len(board[0]) and board[nr][nc] == 'M':
                    mine_count += 1

            if mine_count == 0:
                board[r][c] = 'B'
                for dr, dc in directions:
                    nr, nc = r + dr, c + dc
                    if 0 <= nr < len(board) and 0 <= nc < len(board[0]):
                        queue.append((nr, nc))
            else:
                board[r][c] = str(mine_count)

        return board

• Time Complexity: O(m * n)—visit each cell once.
• Space Complexity: O(m * n)—queue size.

It’s a queued Minesweeper revealer!

Comparing the Two Solutions

• DFS (Best):
• Pros: O(m n), O(m n), intuitive.
• Cons: Recursion stack.
• BFS (Alternative):
• Pros: O(m n), O(m n), systematic.
• Cons: Queue overhead.

DFS wins for simplicity!

Additional Examples and Edge Cases

• [["M"]], [0,0]: [["X"]].
• [["E"]], [0,0]: [["B"]].
• Complex grid: Matches rules.

DFS handles them all!

Complexity Recap

• DFS: Time O(m n), Space O(m n).
• BFS: Time O(m n), Space O(m n).

DFS’s the natural champ!

Key Takeaways

• DFS: Recursive reveal—learn at Python Basics!
• BFS: Queue power.
• Grids: Minesweeper fun.
• Python: Recursion or queues, your pick!

Final Thoughts: Sweep That Board!

LeetCode 529: Minesweeper in Python is a delightful grid challenge. DFS with recursive reveal is your fast track, while BFS offers a queued alternative. Want more? Try LeetCode 200: Number of Islands or LeetCode 994: Rotting Oranges. Ready to click? Head to Solve LeetCode 529 on LeetCode and reveal that board today!