<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 defaultdict
class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        ans=[0]
        hasPathToBoundary=set()
        seen=set()
        def isValid(row,col):
            return 0 <= row < len(grid) and 0 <= col < len(grid[0])

        def dfs(row,col):
            seen.add((row,col))
            directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
            nums=0
            for dx,dy in directions:
                newrow = row + dx
                newcol = col + dy
                if not isValid(newrow,newcol):
                    hasPathToBoundary.add((row,col))
                    return 0
                elif (newrow,newcol) in hasPathToBoundary:
                    hasPathToBoundary.add((row,col))
                    return 0
                elif (newrow,newcol) not in seen and grid[newrow][newcol] == 1:
                    childNum = dfs(newrow,newcol)
                    if childNum == 0:
                        hasPathToBoundary.add((row,col))
                        return 0
                    nums += childNum
            nums += 1
            return nums


        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1 and (i,j) not in seen:
                   ans[0] += dfs(i,j)
        return ans[0]
            


        
48090