Surrounded regions

/*
Logic:This is proper DFS problem.
observation:Only that bulk of O's which is on the corner will be remained O.
so we run dfs for every corner zero and change it to small o that means it will remain
We are changing it into small o so that we can differentiate between which O's to be
O and which to be changed.
O after all operations.
rest all O will be captured.

m-2: dont check the whole matrix, check the corner rows and columns only
*/

from typing import List
def dfs(board, i, j, r, c):
    row = [-1, 0, 1, 0]
    col = [0, 1, 0, -1]

    board[i][j] = 'o'   # mark safe

    for k in range(4):
        new_i = i + row[k]
        new_j = j + col[k]

        if (0 <= new_i < r and
            0 <= new_j < c and
            board[new_i][new_j] == 'O'):

            dfs(board, new_i, new_j, r, c)


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

        r = len(board)
        c = len(board[0])

        # Traverse whole board and check boundary cells
        for i in range(r):
            for j in range(c):
                if (i == 0 or i == r-1 or j == 0 or j == c-1) and board[i][j] == 'O':
                    dfs(board, i, j, r, c)

        # Final conversion
        for i in range(r):
            for j in range(c):
                if board[i][j] == 'o':
                    board[i][j] = 'O'
                else:
                    board[i][j] = 'X'