<p>You are given an <code>m x n</code> binary matrix <code>grid</code>, where <code>0</code> represents a sea cell and <code>1</code> represents a land cell.</p>

<p>A <strong>move</strong> consists of walking from one land cell to another adjacent (<strong>4-directionally</strong>) land cell or walking off the boundary of the <code>grid</code>.</p>

<p>Return <em>the number of land cells in</em> <code>grid</code> <em>for which we cannot walk off the boundary of the grid in any number of <strong>moves</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/enclaves1.jpg" style="width: 333px; height: 333px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three 1s that are enclosed by 0s, and one 1 that is not enclosed because its on the boundary.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/enclaves2.jpg" style="width: 333px; height: 333px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All 1s are either on the boundary or can reach the boundary.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

from collections import deque


class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        directions = [(-1, 0), (1, 0), (0, 1), (0, -1)]

        def in_grid(row, col):
            return 0 <= row < m and 0 <= col < n

        graph = {(-1, -1): set()}
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    graph[(i, j)] = set()

        for row, col in graph:
            for d_row, d_col in directions:
                next_row, next_col = row + d_row, col + d_col
                if not in_grid(next_row, next_col):
                    graph[(-1, -1)].add((row, col))
                    graph[(row, col)].add((-1, -1))
                elif grid[next_row][next_col] == 1:
                    graph[(row, col)].add((next_row, next_col))
                    graph[(next_row, next_col)].add((row, col))

        q = deque([(-1, -1)])
        visited = set([(-1, -1)])
        while q:
            node = q.popleft()
            for neighbor in graph[node]:
                if neighbor not in visited:
                    visited.add(neighbor)
                    q.append(neighbor)

        return len(graph) - len(visited)

        # # conceptually: total size of all islands that don't touch the boundary
        # m, n = len(grid), len(grid[0])
        # directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]

        # def in_grid(row, col):
        #     return 0 <= row < m and 0 <= col < n

        # def bfs(row, col):
        #     on_boundary = False
        #     grid[row][col] = 0
        #     q = deque([(row, col)])
        #     size = 1
        #     while q:
        #         row, col = q.popleft()
        #         for d_row, d_col in directions:
        #             next_row, next_col = row + d_row, col + d_col
        #             if not in_grid(next_row, next_col):
        #                 on_boundary = True
        #             elif grid[next_row][next_col] == 1:
        #                 grid[next_row][next_col] = 0
        #                 q.append((next_row, next_col))
        #                 size += 1
        #     return 0 if on_boundary else size

        # total_size = 0
        # for i in range(m):
        #     for j in range(n):
        #         if grid[i][j] == 1:
        #             total_size += bfs(i, j)
        # return total_size
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