<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>

class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        can_walk_off = set()
        self.sea = set()
        land = set()
        row_l, col_l = len(grid), len(grid[0])
        for r_ind, row in enumerate(grid):
            for col_ind, val in enumerate(row):
                if val == 1:
                    land.add((r_ind, col_ind))
                    if r_ind == 0 or r_ind == row_l-1 or col_ind == 0 or col_ind == col_l-1:
                        can_walk_off.add((r_ind, col_ind))
                    else:
                        print(r_ind, col_ind, row_l, col_l)
                else:#probably don't actually need to keep track of these, and probably faster to array index + less mem
                    self.sea.add((r_ind, col_ind))
        self.seen = set([i for i in can_walk_off])
        stack = [i for i in can_walk_off]
        self.row_l, self.col_l = row_l, col_l
        while stack:
            x, y = stack.pop()
            new_moves = [(x-1, y), (x+1, y), (x, y-1), (x, y+1)]
            for x1, y1 in new_moves:
                if self.is_valid_move((x1,y1)):
                    stack.append((x1,y1))
                    self.seen.add((x1,y1))  
                if (x1,y1) in can_walk_off:
                    can_walk_off.add((x,y))
                    self.seen.add((x,y))
        return len(land) - len(can_walk_off)
            
    
    def is_valid_move(self, i):
        x, y = i
        if x >= 0 and x < self.row_l and y >= 0 and y < self.col_l and (x,y) not in self.seen and (x,y) not in self.sea:
            return True
        return False
    
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