130. Surrounded Regions

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.

// DFS
void helper(vector<vector<char> >& board, int i, int j) {
    if (i < 0 || i >= board.size() || j < 0 || j >= board[0].size() || board[i][j] != 'O') return;
    board[i][j] = '#';
    helper(board, i - 1, j);
    helper(board, i + 1, j);
    helper(board, i, j - 1);
    helper(board, i, j + 1);
}
void solve(vector<vector<char>>& board) { // time: O(m*n); space:O (m*n)
    if (board.empty() || board[0].empty()) return;
    const int m = board.size(), n = board[0].size();
    // Mark any 'O' connected to border 'O'
    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) && board[i][j] == 'O') helper(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';
            else if (board[i][j] == '#') board[i][j] = 'O';
        }
    }
}

// BFS
void solve(vector<vector<char>>& board) { // time: O(m*n); space:O (m*n)
    if (board.empty() || board[0].empty()) return;
    int m = board.size(), n = board[0].size();
    // Mark any 'O' connected to border 'O'
    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) continue;
            if (board[i][j] != 'O') continue;
            board[i][j] = '$';
            queue<int> q{{i * n + j}};
            while (!q.empty()) {
                int t = q.front(), x = t / n, y = t % n; q.pop();
                if (x >= 1 && board[x - 1][y] == 'O') {board[x - 1][y] = '$'; q.push(t - n);}
                if (x < m - 1 && board[x + 1][y] == 'O') {board[x + 1][y] = '$'; q.push(t + n);}
                if (y >= 1 && board[x][y - 1] == 'O') {board[x][y - 1] = '$'; q.push(t - 1);}
                if (y < n - 1 && board[x][y + 1] == 'O') {board[x][y + 1] = '$'; q.push(t + 1);}
            }
        }
    }
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (board[i][j] == 'O') board[i][j] = 'X';
            if (board[i][j] == '$') board[i][j] = 'O';
        }
    }
}

// BFS
void solve(vector<vector<char>>& board) { // time: O(m * n); space:O (m * n)
    if (board.empty() || board[0].empty()) return;
    const int m = board.size(), n = board[0].size();
    const vector<pair<int, int> > dirs({ {-1, 0}, {1, 0}, {0, -1}, {0, 1} });
    // Mark all 'O's connected to border 'O's to different symbol
    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) || board[i][j] != 'O') continue;
            board[i][j] = '$';
            queue<pair<int, int> > q({{i, j}});
            while (!q.empty()) {
                int x = q.front().first, y = q.front().second;
                q.pop();
                for (const pair<int, int>& dir : dirs) {
                    int new_x = x + dir.first, new_y = y + dir.second;
                    if (new_x < 0 || new_x >= m || new_y < 0 || new_y >= n || board[new_x][new_y] != 'O') continue;
                    board[new_x][new_y] = '$';
                    q.push({new_x, new_y});
                }
            }
        }
    }
    // Capture all 'O's inside and change symbol of border 'O's back to 'O'
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (board[i][j] == 'O') board[i][j] = 'X';
            else if (board[i][j] == '$') board[i][j] = 'O';
        }
    }
}