Bfs

Challenge I came across this challenge on leetcode that the goal is to generate a new board after a row, column is selected. Here is an online version of the game. It’s a very BFS like challenge, down below the code. Solution class Solution: def is_valid_coord(self, board, row, col): if row >= 0 and row < len(board) and col >= 0 and col < len(board[0]): return True return False def adjacent_mine(self, board, direction, x, y): adjacent_mine = 0 for nx, ny in direction: new_row = nx + x new_col = ny + y if self.is_valid_coord(board, new_row, new_col) and board[new_row][new_col] == 'M': adjacent_mine += 1 return adjacent_mine def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]: pos_x, pos_y = click if board[pos_x][pos_y] == 'M': board[pos_x][pos_y] = "X" return board queue = [click] # left down right up direction = [[0,-1], [1,-1], [1,0], [1,1], [0,1], [-1,1], [-1,0], [-1,-1]] seen = set() while queue: x, y = queue.pop(0) if board[x][y] == "E" and self.adjacent_mine(board, direction, x, y) == 0: board[x][y] = 'B' seen.add((x, y)) for nx, ny in direction: if self.is_valid_coord(board, nx+x, ny+y) and (nx+x, ny+y) not in seen: queue.append([nx+x, ny+y]) elif board[x][y] == "E" and self.adjacent_mine(board, direction, x, y) >= 0: board[x][y] = str(self.adjacent_mine(board, direction, x, y)) return board Time Complexity: O(m*n) Space Complexity: O(m+n) ...