Surrounded Regions LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Surrounded Regions 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'.

Surrounded Regions Solutions

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'

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