Leetcode

Surrounded Regions

Approach 1: BFS

  • Time:O(mn)
  • Space:O(1)

C++

class Solution {
 public:
  void solve(vector<vector<char>>& board) {
    if (board.empty())
      return;

    const int m = board.size();
    const int n = board[0].size();
    const vector<int> dirs{0, 1, 0, -1, 0};
    queue<pair<int, int>> q;

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (i * j == 0 || i == m - 1 || j == n - 1)
          if (board[i][j] == 'O') {
            q.emplace(i, j);
            board[i][j] = '*';
          }

    // mark grids that stretch from four sides with '*'
    while (!q.empty()) {
      const auto [i, j] = q.front();
      q.pop();
      for (int k = 0; k < 4; ++k) {
        const int x = i + dirs[k];
        const int y = j + dirs[k + 1];
        if (x < 0 || x == m || y < 0 || y == n)
          continue;
        if (board[x][y] != 'O')
          continue;
        q.emplace(x, y);
        board[x][y] = '*';
      }
    }

    for (vector<char>& row : board)
      for (char& c : row)
        if (c == '*')
          c = 'O';
        else if (c == 'O')
          c = 'X';
  }
};

JAVA

class Solution {
  public void solve(char[][] board) {
    if (board.length == 0)
      return;

    final int m = board.length;
    final int n = board[0].length;
    final int[] dirs = {0, 1, 0, -1, 0};
    Queue<int[]> q = new ArrayDeque<>();

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (i * j == 0 || i == m - 1 || j == n - 1)
          if (board[i][j] == 'O') {
            q.offer(new int[] {i, j});
            board[i][j] = '*';
          }

    // mark grids that stretch from four sides with '*'
    while (!q.isEmpty()) {
      final int i = q.peek()[0];
      final int j = q.poll()[1];
      for (int k = 0; k < 4; ++k) {
        final int x = i + dirs[k];
        final int y = j + dirs[k + 1];
        if (x < 0 || x == m || y < 0 || y == n)
          continue;
        if (board[x][y] != 'O')
          continue;
        q.offer(new int[] {x, y});
        board[x][y] = '*';
      }
    }

    for (char[] row : board)
      for (int i = 0; i < row.length; ++i)
        if (row[i] == '*')
          row[i] = 'O';
        else if (row[i] == 'O')
          row[i] = 'X';
  }
}

Python

class Solution:
  def solve(self, board: List[List[str]]) -> None:
    if not board:
      return

    m = len(board)
    n = len(board[0])
    dirs = [0, 1, 0, -1, 0]
    q = deque()

    for i in range(m):
      for j in range(n):
        if i * j == 0 or i == m - 1 or j == n - 1:
          if board[i][j] == 'O':
            q.append((i, j))
            board[i][j] = '*'

    # mark grids that stretch from four sides with '*'
    while q:
      i, j = q.popleft()
      for k in range(4):
        x = i + dirs[k]
        y = j + dirs[k + 1]
        if x < 0 or x == m or y < 0 or y == n:
          continue
        if board[x][y] != 'O':
          continue
        q.append((x, y))
        board[x][y] = '*'

    for row in board:
      for i, c in enumerate(row):
        row[i] = 'O' if c == '*' else 'X'

Approach 2: DFS

  • Time:O(mn)
  • Space:O(1)

C++

class Solution {
 public:
  void solve(vector<vector<char>>& board) {
    if (board.empty())
      return;

    const int m = board.size();
    const int n = board[0].size();

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (i * j == 0 || i == m - 1 || j == n - 1)
          dfs(board, i, j);

    for (vector<char>& row : board)
      for (char& c : row)
        if (c == '*')
          c = 'O';
        else if (c == 'O')
          c = 'X';
  }

 private:
  // mark 'O' grids that stretch from four sides with '*'
  void dfs(vector<vector<char>>& board, int i, int j) {
    if (i < 0 || i == board.size() || j < 0 || j == board[0].size())
      return;
    if (board[i][j] != 'O')
      return;

    board[i][j] = '*';
    dfs(board, i + 1, j);
    dfs(board, i - 1, j);
    dfs(board, i, j + 1);
    dfs(board, i, j - 1);
  }
};

JAVA

class Solution {
  public void solve(char[][] board) {
    if (board.length == 0)
      return;

    final int m = board.length;
    final int n = board[0].length;

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (i * j == 0 || i == m - 1 || j == n - 1)
          dfs(board, i, j);

    for (char[] row : board)
      for (int i = 0; i < row.length; ++i)
        if (row[i] == '*')
          row[i] = 'O';
        else if (row[i] == 'O')
          row[i] = 'X';
  }

  // mark 'O' grids that stretch from four sides with '*'
  private void dfs(char[][] board, int i, int j) {
    if (i < 0 || i == board.length || j < 0 || j == board[0].length)
      return;
    if (board[i][j] != 'O')
      return;

    board[i][j] = '*';
    dfs(board, i + 1, j);
    dfs(board, i - 1, j);
    dfs(board, i, j + 1);
    dfs(board, i, j - 1);
  }
}

Python

class Solution:
  def solve(self, board: List[List[str]]) -> None:
    if not board:
      return

    m = len(board)
    n = len(board[0])

    def dfs(i: int, j: int) -> None:
      if i < 0 or i == m or j < 0 or j == n:
        return
      if board[i][j] != 'O':
        return

      board[i][j] = '*'
      dfs(i + 1, j)
      dfs(i - 1, j)
      dfs(i, j + 1)
      dfs(i, j - 1)

    for i in range(m):
      for j in range(n):
        if i * j == 0 or i == m - 1 or j == n - 1:
          dfs(i, j)

    for row in board:
      for i, c in enumerate(row):
        row[i] = 'O' if c == '*' else 'X'