Solution: Number of Closed Islands

from collections import deque

class Solution:
  def countClosedIslands(self, matrix):
    countClosedIslands = 0
    
    self.n = len(matrix)
    self.m = len(matrix[0])

    self.matrix = matrix

    self.visited = [[False for y in range(self.m)] for x in range(self.n)]

    for i in range(self.n):
      for j in range(self.m):
        if self.matrix[i][j] == 1 and not self.visited[i][j]:
          is_closed = self.bfs(i, j)
          if is_closed:
            countClosedIslands += 1

    return countClosedIslands

  def bfs(self, p, q):
    queue = deque([(p, q)])

    closed = True

    while queue:
      row, col = queue.popleft()

      if row < 0 or row >= self.n or col < 0 or col >= self.m:
        closed = False
        continue

      if self.matrix[row][col] == 0:

from collections import deque

class Solution:
  def countClosedIslands(self, matrix):
    countClosedIslands = 0
    # TODO: Write your code here
    # during the visit, we want to check if any node of island reside on the edge, then force return 0 if it is
    # Time complexity: O(row * col)
    # Space complexity: O(row * col) 
    for row in range(len(matrix)):
      for col in range(len(matrix[0])):
        if matrix[row][col] == 1:
          countClosedIslands += self.visitIsland(row, col, matrix)
    return countClosedIslands
  def visitIsland(self, row, col, matrix):
    num_return = 1
    neighbors = deque([(row, col)])
    while neighbors:
      row, col = neighbors.popleft()
      if row < 0 or row >= len(matrix) or col < 0 or col >= len(matrix[0]):
        continue