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.
Related Topics:
Depth-first Search, Breadth-first Search, Union Find
Similar Questions:
// OJ: https://leetcode.com/problems/surrounded-regions/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
int M, N, dirs[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
void dfs(vector<vector<char>> &board, int x, int y) {
if (x < 0 || x >= M || y < 0 || y >= N || board[x][y] != 'O') return;
board[x][y] = '#';
for (auto &dir : dirs) dfs(board, x + dir[0], y + dir[1]);
}
public:
void solve(vector<vector<char>>& board) {
if (board.empty() || board[0].empty()) return;
M = board.size(), N = board[0].size();
for (int i = 0; i < M; ++i) {
dfs(board, i, 0);
dfs(board, i, N - 1);
}
for (int j = 0; j < N; ++j) {
dfs(board, 0, j);
dfs(board, M - 1, j);
}
for (auto &row : board) {
for (auto &cell : row) {
cell = cell == '#' ? 'O' : 'X';
}
}
}
};// OJ: https://leetcode.com/problems/surrounded-regions
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
#define FILL(x, y) do { if (board[x][y] == 'O') { q.emplace(x, y); board[x][y] = '#'; }} while(0)
class Solution {
private:
const int dirs[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
public:
void solve(vector<vector<char>>& board) {
if (board.empty() || board[0].empty()) return;
int M = board.size(), N = board[0].size();
queue<pair<int, int>> q;
for (int i = 0; i < N; ++i) {
FILL(0, i);
FILL(M - 1, i);
}
for (int i = 1; i < M - 1; ++i) {
FILL(i, 0);
FILL(i, N - 1);
}
while (q.size()) {
auto p = q.front();
q.pop();
for (auto &dir : dirs) {
int x = p.first + dir[0], y = p.second + dir[1];
if (x >= 0 && x < M && y >= 0 && y < N) FILL(x, y);
}
}
for (auto &row : board)
for (char &c : row) c = c == '#' ? 'O' : 'X';
}
};// OJ: https://leetcode.com/problems/surrounded-regions
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
// Ref: https://discuss.leetcode.com/topic/1944/solve-it-using-union-find
class UnionFind {
vector<int> id, rank;
public:
UnionFind(int n): id(n), rank(n, 1) {
for (int i = 0; i < n; ++i) id[i] = i;
}
void connect(int a, int b) {
int x = find(a), y = find(b);
if (x == y) return;
if (rank[x] <= rank[y]) {
id[x] = y;
if (rank[x] == rank[y]) rank[y]++;
} else id[y] = x;
}
int find(int a) {
return id[a] == a ? a : (id[a] = find(id[a]));
}
bool connected(int a, int b) {
return find(a) == find(b);
}
};
class Solution {
int M, N, dirs[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
inline int get(int x, int y) { return N * x + y; }
public:
void solve(vector<vector<char>>& board) {
if (board.empty() || board[0].empty()) return;
M = board.size(), N = board[0].size();
UnionFind uf(M * N + 1);
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (board[i][j] != 'O') continue;
if (!i || i == M - 1 || !j || j == N - 1) uf.connect(get(i, j), M * N);
for (auto &dir : dirs) {
int x = i + dir[0], y = j + dir[1];
if (x < 0 || x >= M || y < 0 || y >= N || board[x][y] != 'O') continue;
uf.connect(get(i, j), get(x, y));
}
}
}
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (!uf.connected(get(i, j), M * N)) board[i][j] = 'X';
}
}
}
};