LeetCode #529 — MEDIUM

Minesweeper

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 where click = [click_r, click_c] represents the next click position among all the unrevealed squares ('M' or 'E').

Return the board after revealing this position according to the following rules:

If a mine 'M' is revealed, then the game is over. You should change it to 'X'.
If an empty square 'E' with no adjacent mines is revealed, then change it to a revealed blank 'B' and all of its adjacent unrevealed squares should be revealed recursively.
If an empty square 'E' with at least one adjacent mine is revealed, then change it to a digit ('1' to '8') representing the number of adjacent mines.
Return the board when no more squares will be revealed.

Example 1:

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","1","1","B"],["B","B","B","B","B"]]

Example 2:

Input: board = [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]], click = [1,2]
Output: [["B","1","E","1","B"],["B","1","X","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]

Constraints:

m == board.length
n == board[i].length
1 <= m, n <= 50
board[i][j] is either 'M', 'E', 'B', or a digit from '1' to '8'.
click.length == 2
0 <= click_r < m
0 <= click_c < n
board[click_r][click_c] is either 'M' or 'E'.

Step 01

Brute Force Baseline

Problem summary: 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 where click = [clickr, clickc] represents the next click position among all the unrevealed squares ('M' or 'E'). Return the board after revealing this position according to the following rules: If a mine 'M' is revealed, then the game is over. You should change it to 'X'. If an empty square 'E' with no adjacent mines is revealed, then change it to a revealed blank 'B' and all of its adjacent

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[["E","E","E","E","E"],["E","E","M","E","E"],["E","E","E","E","E"],["E","E","E","E","E"]]
[3,0]

Example 2

[["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
[1,2]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #529: Minesweeper
class Solution {
    private char[][] board;
    private int m;
    private int n;

    public char[][] updateBoard(char[][] board, int[] click) {
        m = board.length;
        n = board[0].length;
        this.board = board;
        int i = click[0], j = click[1];
        if (board[i][j] == 'M') {
            board[i][j] = 'X';
        } else {
            dfs(i, j);
        }
        return board;
    }

    private void dfs(int i, int j) {
        int cnt = 0;
        for (int x = i - 1; x <= i + 1; ++x) {
            for (int y = j - 1; y <= j + 1; ++y) {
                if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M') {
                    ++cnt;
                }
            }
        }
        if (cnt > 0) {
            board[i][j] = (char) (cnt + '0');
        } else {
            board[i][j] = 'B';
            for (int x = i - 1; x <= i + 1; ++x) {
                for (int y = j - 1; y <= j + 1; ++y) {
                    if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E') {
                        dfs(x, y);
                    }
                }
            }
        }
    }
}

// Accepted solution for LeetCode #529: Minesweeper
func updateBoard(board [][]byte, click []int) [][]byte {
	m, n := len(board), len(board[0])
	i, j := click[0], click[1]

	var dfs func(i, j int)
	dfs = func(i, j int) {
		cnt := 0
		for x := i - 1; x <= i+1; x++ {
			for y := j - 1; y <= j+1; y++ {
				if x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M' {
					cnt++
				}
			}
		}
		if cnt > 0 {
			board[i][j] = byte(cnt + '0')
			return
		}
		board[i][j] = 'B'
		for x := i - 1; x <= i+1; x++ {
			for y := j - 1; y <= j+1; y++ {
				if x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E' {
					dfs(x, y)
				}
			}
		}
	}

	if board[i][j] == 'M' {
		board[i][j] = 'X'
	} else {
		dfs(i, j)
	}
	return board
}

# Accepted solution for LeetCode #529: Minesweeper
class Solution:
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        def dfs(i: int, j: int):
            cnt = 0
            for x in range(i - 1, i + 2):
                for y in range(j - 1, j + 2):
                    if 0 <= x < m and 0 <= y < n and board[x][y] == "M":
                        cnt += 1
            if cnt:
                board[i][j] = str(cnt)
            else:
                board[i][j] = "B"
                for x in range(i - 1, i + 2):
                    for y in range(j - 1, j + 2):
                        if 0 <= x < m and 0 <= y < n and board[x][y] == "E":
                            dfs(x, y)

        m, n = len(board), len(board[0])
        i, j = click
        if board[i][j] == "M":
            board[i][j] = "X"
        else:
            dfs(i, j)
        return board

// Accepted solution for LeetCode #529: Minesweeper
struct Solution;

use std::collections::VecDeque;

struct Point {
    i: usize,
    j: usize,
}

macro_rules! point {
    ($i:expr,$j:expr) => {
        Point { i: $i, j: $j }
    };
}

impl Point {
    fn adj(&self, n: usize, m: usize) -> Vec<Point> {
        let mut res: Vec<Point> = vec![];
        for i in -1..=1 {
            for j in -1..=1 {
                let r = self.i as i32 + i;
                let c = self.j as i32 + j;
                if r < 0 || r > (n - 1) as i32 || c < 0 || c > (m - 1) as i32 {
                    continue;
                }
                res.push(point!(r as usize, c as usize))
            }
        }
        res
    }
}

impl Solution {
    fn update_board(mut board: Vec<Vec<char>>, click: Vec<i32>) -> Vec<Vec<char>> {
        let n = board.len();
        let m = board[0].len();
        let i = click[0] as usize;
        let j = click[1] as usize;
        if board[i][j] == 'M' {
            board[i][j] = 'X';
            return board;
        }
        let mut visited: Vec<Vec<bool>> = vec![vec![false; m]; n];
        let mut queue: VecDeque<Point> = VecDeque::new();
        queue.push_back(Point { i, j });
        while let Some(p) = queue.pop_front() {
            visited[p.i][p.j] = true;
            if board[p.i][p.j] == 'E' {
                let mut sum = 0;
                for q in p.adj(n, m) {
                    if board[q.i][q.j] == 'M' {
                        sum += 1;
                    }
                }
                if sum == 0 {
                    board[p.i][p.j] = 'B';
                    for q in p.adj(n, m) {
                        if !visited[q.i][q.j] {
                            queue.push_back(q);
                        }
                    }
                } else {
                    board[p.i][p.j] = (b'0' + sum) as char;
                }
            }
        }
        board
    }
}

#[test]
fn test() {
    let board = vec_vec_char![
        ['E', 'E', 'E', 'E', 'E'],
        ['E', 'E', 'M', 'E', 'E'],
        ['E', 'E', 'E', 'E', 'E'],
        ['E', 'E', 'E', 'E', 'E']
    ];
    let click = vec![3, 0];
    let res = vec_vec_char![
        ['B', '1', 'E', '1', 'B'],
        ['B', '1', 'M', '1', 'B'],
        ['B', '1', '1', '1', 'B'],
        ['B', 'B', 'B', 'B', 'B']
    ];
    assert_eq!(Solution::update_board(board, click), res);
}

// Accepted solution for LeetCode #529: Minesweeper
function updateBoard(board: string[][], click: number[]): string[][] {
    const m = board.length;
    const n = board[0].length;
    const [i, j] = click;

    const dfs = (i: number, j: number) => {
        let cnt = 0;
        for (let x = i - 1; x <= i + 1; ++x) {
            for (let y = j - 1; y <= j + 1; ++y) {
                if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] === 'M') {
                    ++cnt;
                }
            }
        }
        if (cnt > 0) {
            board[i][j] = cnt.toString();
            return;
        }
        board[i][j] = 'B';
        for (let x = i - 1; x <= i + 1; ++x) {
            for (let y = j - 1; y <= j + 1; ++y) {
                if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] === 'E') {
                    dfs(x, y);
                }
            }
        }
    };

    if (board[i][j] === 'M') {
        board[i][j] = 'X';
    } else {
        dfs(i, j);
    }
    return board;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.