130

130. Surrounded Regions (Medium)

Given an m x n matrix board containing 'X' and 'O', capture all regions that are 4-directionally 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'.

Companies:
Google, Amazon, Microsoft

Related Topics:
Array, Depth-First Search, Breadth-First Search, Union Find, Matrix

Similar Questions:

• Number of Islands (Medium)
• Walls and Gates (Medium)

Solution 1. DFS

// OJ: https://leetcode.com/problems/surrounded-regions/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    void solve(vector<vector<char>>& A) {
        int M = A.size(), N = A[0].size(), dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
        function<void(int, int)> dfs = [&](int x, int y) {
            if (x < 0 || x >= M || y < 0 || y >= N || A[x][y] != 'O') return;
            A[x][y] = '#';
            for (auto &[dx, dy] : dirs) dfs(x + dx, y + dy);
        };
        for (int i = 0; i < M; ++i) dfs(i, 0), dfs(i, N - 1);
        for (int j = 0; j < N; ++j) dfs(0, j), dfs(M - 1, j);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                A[i][j] = A[i][j] == '#' ? 'O' : 'X';
            }
        }
    }
};

Solution 2. BFS

// OJ: https://leetcode.com/problems/surrounded-regions
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    void solve(vector<vector<char>>& A) {
        int M = A.size(), N = A[0].size(), dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
        auto bfs = [&](int x, int y) {
            if (A[x][y] != 'O') return;
            queue<pair<int, int>> q{{{x, y}}};
            A[x][y] = '#';
            while (q.size()) {
                auto [x, y] = q.front();
                q.pop();
                for (auto &[dx, dy] : dirs) {
                    int a = x + dx, b = y + dy;
                    if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] != 'O') continue;
                    A[a][b] = '#';
                    q.emplace(a, b);
                }
            }
        };
        for (int i = 0; i < M; ++i) bfs(i, 0), bfs(i, N - 1);
        for (int j = 0; j < N; ++j) bfs(0, j), bfs(M - 1, j);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                A[i][j] = A[i][j] == '#' ? 'O' : 'X';
            }
        }
    }
};

Solution 3. Union Find

// OJ: https://leetcode.com/problems/surrounded-regions
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class UnionFind {
    vector<int> id;
public:
    UnionFind(int N) : id(N) {
        iota(begin(id), end(id), 0);
    }
    int find(int a) {
        return id[a] == a ? a : (id[a] = find(id[a]));
    }
    void connect(int a, int b) {
        id[find(a)] = find(b);
    }
    bool connected(int a, int b) {
        return find(a) == find(b);
    }
};
class Solution {
public:
    void solve(vector<vector<char>>& A) {
        int M = A.size(), N = A[0].size(), dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}}, edge = M * N;
        auto key = [&](int x, int y) { return x * N + y; };
        UnionFind uf(M * N + 1);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] != 'O') continue;
                if (i == 0 || i == M - 1 || j == 0 || j == N - 1) uf.connect(key(i, j), edge);
                for (auto &[dx, dy] : dirs) {
                    int a = i + dx, b = j + dy;
                    if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] != 'O') continue;
                    uf.connect(key(a, b), key(i, j));
                }
            }
        }
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                A[i][j] = uf.connected(key(i, j), edge) ? 'O' : 'X';
            }
        }
    }
};