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.lengthn == board[i].length1 <= m, n <= 50board[i][j]is either'M','E','B', or a digit from'1'to'8'.click.length == 20 <= clickr < m0 <= clickc < nboard[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
Which of the following is a good use case for backtracking?
What's the output of running the following function using the following tree as input?
1def serialize(root):
2 res = []
3 def dfs(root):
4 if not root:
5 res.append('x')
6 return
7 res.append(root.val)
8 dfs(root.left)
9 dfs(root.right)
10 dfs(root)
11 return ' '.join(res)
121import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4 StringJoiner res = new StringJoiner(" ");
5 serializeDFS(root, res);
6 return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10 if (root == null) {
11 result.add("x");
12 return;
13 }
14 result.add(Integer.toString(root.val));
15 serializeDFS(root.left, result);
16 serializeDFS(root.right, result);
17}
181function serialize(root) {
2 let res = [];
3 serialize_dfs(root, res);
4 return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8 if (!root) {
9 res.push("x");
10 return;
11 }
12 res.push(root.val);
13 serialize_dfs(root.left, res);
14 serialize_dfs(root.right, res);
15}
16Solution Implementation
Which of the following problems can be solved with backtracking (select multiple)
Consider the classic dynamic programming of longest increasing subsequence:
Find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.
For example, the length of LIS for [50, 3, 10, 7, 40, 80] is 4 and LIS is
[3, 7, 40, 80].
What is the recurrence relation?
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