130. Surrounded Regions

Updated: 2024-03-12
2 min read

On This Page

LeetCode problem

Naive Solution:

A naive solution would be to iterate through each cell in the grid, and for each O, check if it is surrounded by X’s in all four directions (up, down, left, and right). If so, flip it to X. However, this method has a high time complexity and does not take advantage of any properties of the problem.

Approach: The more efficient solution is to perform a Depth-First Search (DFS) starting from the border O’s.

DFS is a way to explore a graph or tree by visiting as deep as possible in a single path before backtracking.


  1. In this problem, we will mark the border O’s and all their adjacent O’s as not to be flipped to X. Will temporary change these cells with O!.
    1. This means that if (later) cell is marked as O! then we will change it back to O. All other cells should be X.
    2. Loop through borders.
      1. If O is in cell then check its neighbors (dfs).
        1. Border cell mark to O!
  2. Then, we can iterate through the entire grid, flipping any O’s that are not marked as not to be flipped.
class Solution:
    def dfs(self, board, row, col):
        # If the current cell is out of bounds or not an 'O', return and stop DFS
        if (
            row < 0
            or col < 0
            or row >= len(board)
            or col >= len(board[0])
            or board[row][col] != "O"  # X or O!

        # Mark the current cell as 'O!' (Don't flip)
        board[row][col] = "O!"

        # Define the possible directions to move (up, down, left, right)
        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]

        # Explore each direction recursively by calling the DFS function
        for dr, dc in directions:
            self.dfs(board, row + dr, col + dc)

    def solve(self, board):
        m = len(board)
        n = len(board[0])

        # Iterate through the border cells
        # rows
        for row in range(m):
            for col in [0, n - 1]:  # left border coll, right border coll
                # If a border cell contains 'O', perform DFS on that cell
                if board[row][col] == "O":
                    self.dfs(board, row, col)
        # cells
        for col in range(n):
            for row in [0, m - 1]:  # upper border row, bottom border row
                if board[row][col] == "O":
                    self.dfs(board, row, col)

        # Iterate through the entire grid
        for row in range(m):
            for col in range(n):
                if board[row][col] == "O!":
                    board[row][col] = "O"
                    board[row][col] = "X"

        return board