Minesweeper
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 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 <= clickr < m
0 <= clickc < n
board[clickr][clickc]
is either'M'
or'E'
.
Solution
There are a few things we need to keep in mind when implementing the problem:
- blank squares (
'B'
) and digit squares ('1'
-'8'
) should be treated differently - a blank square's neighbours need to be revealed recursively
- only recursively reveal empty squares ('E')
We are given the click
coordinate to update the board. If click
is a mine ('M'
), then we change it to 'X'
.
If it's an empty square, then the update
function updates the empty square recursively.
Implementation
1def updateBoard(board: List[List[str]], click: List[int]) -> List[List[str]]:
2 if board == []: return []
3 row = len(board)
4 col = len(board[0])
5
6 def isbomb(i, j):
7 if i < 0 or j < 0 or i >= row or j >= col:
8 return False
9 elif board[i][j] == 'M' or board[i][j] == 'X':
10 return True
11 else: return False
12
13 def update(r, c):
14 if r < 0 or c < 0 or r >= row or c >= col: return
15 if board[r][c] == 'E': # empty square
16 bombs = 0 # number of neighbouring bombs
17 for i in range(r-1, r+2):
18 for j in range(c-1, c+2):
19 if isbomb(i, j): bombs += 1
20 if bombs: # update cell with number of bombs
21 board[r][c] = chr(48 + bombs)
22 else: # blank cell (0 bombs)
23 board[r][c] = 'B'
24 for i in range(r-1, r+2): # update neighbours
25 for j in range(c-1, c+2):
26 update(i, j)
27 # else already revealed, do nothing
28
29 r, c = click[0], click[1]
30 if board[r][c] == 'M': # bomb
31 board[r][c] = 'X'
32 else: # empty square, or revealed square
33 update(r, c)
34 return board
In a binary min heap, the minimum element can be found in:
What is the space complexity of the following code?
1int sum(int n) {
2 if (n <= 0) {
3 return 0;
4 }
5 return n + sum(n - 1);
6}
What are the two properties the problem needs to have for dynamic programming to be applicable? (Select 2)
Which algorithm should you use to find a node that is close to the root of the tree?
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