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Abhishek Chaudhary

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# Number of Enclaves

You are given an `m x n` binary matrix `grid`, where `0` represents a sea cell and `1` represents a land cell.

A move consists of walking from one land cell to another adjacent (4-directionally) land cell or walking off the boundary of the `grid`.

Return the number of land cells in `grid` for which we cannot walk off the boundary of the grid in any number of moves.

Example 1:

Input: grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
Output: 3
Explanation: There are three 1s that are enclosed by 0s, and one 1 that is not enclosed because its on the boundary.

Example 2:

Input: grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
Output: 0
Explanation: All 1s are either on the boundary or can reach the boundary.

Constraints:

• `m == grid.length`
• `n == grid[i].length`
• `1 <= m, n <= 500`
• `grid[i][j]` is either `0` or `1`.

SOLUTION:

``````class Solution:
def dfs(self, grid, i, j, m, n, visited):
grid[i][j] = 0
for x, y in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]:
if 0 <= x < m and 0 <= y < n and grid[x][y] == 1 and (x, y) not in visited:
self.dfs(grid, x, y, m, n, visited)

def numEnclaves(self, grid: List[List[int]]) -> int:
m = len(grid)
n = len(grid[0])
boundary = set()
for i in range(m):
for i in range(n):