0130. Surrounded Regions

Given an m x n matrix board containing 'X' and 'O', capture all regions surrounded by 'X'.

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

Example 1:

Input: board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
Output: [["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
Explanation: Surrounded regions should not 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.

Example 2:

Input: board = [["X"]]
Output: [["X"]]

Constraints:

• m == board.length
• n == board[i].length
• 1 <= m, n <= 200
• board[i][j] is 'X' or 'O'.

Analysis

We cannot just find all 'O' that isn't at the boarder and expand through dfs, because consider the case:

X X X X
X O O X
X O O X
X O X X

This case will cause all the 'O' not be able to be flipped, so our first step is to find all the 'O' that is on the boarder and mark all the connected 'O' as visited (either allocate a new set or special symbol other than 'O' or 'X'). This will takes $O(n)$ time at worst. Step 2 would be find all the 'O' remaining in the board and flip them.

• Time: $O(N)$
• Space: $O(N)$ if cannot modify the original board, or $O(1)$

Code

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        if (board.empty() || board[0].empty()) return;
        int m = board.size(), n = board[0].size();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    if (board[i][j] == 'O') dfs(board, i , j);
                }
            }   
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                // recover
                if (board[i][j] == '$') board[i][j] = 'O';
            }
        }
    }

    // find all 'O' that are invalid -- from boarder
    void dfs(vector<vector<char>> &board, int x, int y) {
        int m = board.size(), n = board[0].size();
        vector<vector<int>> dir{{0,-1},{-1,0},{0,1},{1,0}};
        board[x][y] = '$';
        for (int i = 0; i < dir.size(); ++i) {
            int dx = x + dir[i][0], dy = y + dir[i][1];
            if (dx >= 0 && dx < m && dy > 0 && dy < n && board[dx][dy] == 'O') {
                dfs(board, dx, dy);
            }
        }
    }
};

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Last update: April 1, 2022