<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

def solve(grid):
    m,n = len(grid),len(grid[0])
    visited = [[0]*n for i in range(m)]
    #boundry land can't be in answer 
    q = deque([])

    # frist row
    for j in range(n):
        if grid[0][j]==1:
            q.append((0,j))
            visited[0][j] = 1

    # last row
    for j in range(n):
        if grid[m-1][j]==1:
            q.append((m-1,j))
            visited[m-1][j] = 1

    #first column
    for i in range(m):
        if grid[i][0]==1:
            q.append((i,0))
            visited[i][0] = 1

    #last column
    for i in range(m):
        if grid[i][n-1]==1:
            q.append((i,n-1))
            visited[i][n-1] = 1

    return bfs(q,grid,visited)
def bfs(q,grid,visited):
    dx = [-1,0,1,0]
    dy = [0,1,0,-1]
    m,n = len(grid),len(grid[0])
    while len(q):
        ix,iy = q.popleft()
        for i in range(4):
            x = ix+dx[i]
            y = iy+dy[i]
            if 0<=x<m and 0<=y<n and grid[x][y]==1 and visited[x][y]==0:
                visited[x][y] = 1
                q.append((x,y))
    return visited


def dfs(i,j,grid,visited):
    visited[i][j] = 1
    dx = [-1,0,1,0]
    dy = [0,1,0,-1]
    m,n = len(grid),len(grid[0])
    for k in range(4):
        xx = i+dx[k]
        yy = j+dy[k]
        if 0<=xx<m and 0<=yy<n and visited[xx][yy]!=1 and grid[xx][yy]==1:
            dfs(xx,yy,grid,visited)



def solve_dfs(grid):
    m,n = len(grid),len(grid[0])
    visited = [[0]*n for i in range(m)]
    # frist row
    for j in range(n):
        if grid[0][j]==1 and visited[0][j]!=1:
            dfs(0,j,grid,visited)

    # last row
    for j in range(n):
        if grid[m-1][j]==1 and visited[m-1][j]!=1:
            dfs(m-1,j,grid,visited)
           

    #first column
    for i in range(m):
        if grid[i][0]==1 and visited[i][0]!=1:
           dfs(i,0,grid,visited)

    #last column
    for i in range(m):
        if grid[i][n-1]==1 and visited[i][n-1]!=1:
           dfs(i,n-1,grid,visited)
    return visited


class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        #can also solved by DFS
        visited = solve_dfs(grid)
        # visited = solve(grid)
        m,n = len(grid),len(grid[0])
        count = 0
        for i in range(m):
            for j in range(n):
                if visited[i][j]==0 and grid[i][j]==1:
                    count+=1
        return count
        
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