leetcode_gym/_problems/surroundedRegion.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Surrounded Region	2020-06-16	union find	https://leetcode.com/problems/surrounded-regions	Attempted

Problem

Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Example:

X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X
Explanation:

Surrounded regions shouldn’t be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically.

Solution

We need to have some way to figure out which cells are "surrounded" by X. If they are surrounded, we don't flip them. To do this:

• First, go thru border cells and run DFS on them, temporarily turning them to another color 'T'.
• Then iterate thru matrix (use itertools.product()) and change to 'X' if not 'T', else 'O

{% highlight python %}

from itertools import product

''' https://leetcode.com/problems/surrounded-regions/discuss/691646/Python-O(mn)-3-colors-dfs-explained '''

board = [ ["X","X","X","X"], ["X","O","O","X"], ["X","X","O","X"], ["X","O","X","X"] ]

class Solution:

def dfs(self, i, j):
    
    if i<0 or j<0 or i>=self.M or j>=self.N or self.board[i][j] != "O":
        return
    self.board[i][j] = 'T'
    neib_list = [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]
    for x, y in neib_list:
        self.dfs(x, y)

def solve(self, board):
    if not board: return 0
    self.board, self.M, self.N = board, len(board), len(board[0])
    
    # run dfs on sides
    for i in range(0, self.M):
        self.dfs(i,0)           # left side
        self.dfs(i,self.N-1)    # right side
    
    # run dfs on top and bottom
    for j in range(0, self.N):
        self.dfs(0,j)           # top
        self.dfs(self.M-1,j)    # bottom
    
    print(board)

    # itertools.product() is equivalent to nested forloop
    for i,j in product(range(self.M), range(self.N)):
        board[i][j] = "X" if board[i][j] != "T" else "O"

sol = Solution() sol.solve(board)

print(board)

{% endhighlight %}