<p>You are given two integers <code>m</code> and <code>n</code> representing a <strong>0-indexed</strong> <code>m x n</code> grid. You are also given two 2D integer arrays <code>guards</code> and <code>walls</code> where <code>guards[i] = [row<sub>i</sub>, col<sub>i</sub>]</code> and <code>walls[j] = [row<sub>j</sub>, col<sub>j</sub>]</code> represent the positions of the <code>i<sup>th</sup></code> guard and <code>j<sup>th</sup></code> wall respectively.</p>

<p>A guard can see <b>every</b> cell in the four cardinal directions (north, east, south, or west) starting from their position unless <strong>obstructed</strong> by a wall or another guard. A cell is <strong>guarded</strong> if there is <strong>at least</strong> one guard that can see it.</p>

<p>Return<em> the number of unoccupied cells that are <strong>not</strong> <strong>guarded</strong>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/10/example1drawio2.png" style="width: 300px; height: 204px;" />
<pre>
<strong>Input:</strong> m = 4, n = 6, guards = [[0,0],[1,1],[2,3]], walls = [[0,1],[2,2],[1,4]]
<strong>Output:</strong> 7
<strong>Explanation:</strong> The guarded and unguarded cells are shown in red and green respectively in the above diagram.
There are a total of 7 unguarded cells, so we return 7.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/10/example2drawio.png" style="width: 200px; height: 201px;" />
<pre>
<strong>Input:</strong> m = 3, n = 3, guards = [[1,1]], walls = [[0,1],[1,0],[2,1],[1,2]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The unguarded cells are shown in green in the above diagram.
There are a total of 4 unguarded cells, so we return 4.
</pre>

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

<ul>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>2 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= guards.length, walls.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>2 &lt;= guards.length + walls.length &lt;= m * n</code></li>
	<li><code>guards[i].length == walls[j].length == 2</code></li>
	<li><code>0 &lt;= row<sub>i</sub>, row<sub>j</sub> &lt; m</code></li>
	<li><code>0 &lt;= col<sub>i</sub>, col<sub>j</sub> &lt; n</code></li>
	<li>All the positions in <code>guards</code> and <code>walls</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def countUnguarded(self, m: int, n: int, guards: List[List[int]], walls: List[List[int]]) -> int:
        # https://github.com/doocs/leetcode/tree/main/solution/2200-2299/2257.Count%20Unguarded%20Cells%20in%20the%20Grid
        safe = set()
        guards = set(tuple(x) for x in guards)
        walls = set(tuple(x) for x in walls)

        for x, y in guards:
            for i, j in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                nx, ny = x + i, y + j
                while 0 <= nx < m and 0 <= ny < n and (nx, ny) not in walls and (nx, ny) not in guards:
                    safe.add((nx, ny))
                    nx += i
                    ny += j
                    
        return m * n - len(safe) - len(guards) - len(walls)

        # safe = set()
        # guards = set([(x[0], x[1]) for x in guards])
        # walls = set([(x[0], x[1]) for x in walls])
        # for x, y in guards:
        #     if y != n-1:
        #         for j in range(y+1, n) :
        #             if (x, j) not in walls and (x, j) not in guards:
        #                 safe.add((x, j))
        #             else:
        #                 break
        #     if y != 0:
        #         for j in range(y-1, -1, -1):
        #             if (x, j) not in walls and (x, j) not in guards:
        #                 safe.add((x, j))
        #             else:
        #                 break
        #     if x != m-1:
        #         for i in range(x+1, m):
        #             if (i, y) not in walls and (i, y) not in guards:
        #                 safe.add((i, y))
        #             else:
        #                 break
        #     if x != 0:
        #         for i in range(x-1, -1, -1):
        #             if (i, y) not in walls and (i, y) not in guards:
        #                 safe.add((i, y))
        #             else:
        #                 break
        # return m*n - len(safe) - len(guards) - len(walls)

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