You are given an
m x n binary matrix
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
Return the number of land cells in
grid for which we cannot walk off the boundary of the grid in any number of moves.
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.
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.
m == grid.length
n == grid[i].length
1 <= m, n <= 500
class Solution: cnt=0 def numEnclaves(self, grid: List[List[int]]) -> int: m, n=len(grid), len(grid) def count(i,j, grid): # if its boundary if i<0 or j<0 or i>=m or j>=n: return -(m*n) if grid[i][j]==0: return 0 # mark visited cell grid[i][j]=0 # check 4-direction cell top=count(i-1,j,grid) bottom=count(i+1,j,grid) left=count(i,j-1,grid) right=count(i,j+1,grid) return 1+ top + bottom + left + right for i in range(m): for j in range(n): if grid[i][j]==1: check=count(i,j,grid) if check>0: self.cnt+=check return self.cnt
1on the edge, do DFS and clean all connected
class Solution: def numEnclaves(self, A: List[List[int]]) -> int: def dfs(i, j): A[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 A[x][y]: dfs(x, y) m, n = len(A), len(A) for i in range(m): for j in range(n): if A[i][j] == 1 and (i == 0 or j == 0 or i == m - 1 or j == n - 1): dfs(i, j) return sum(sum(row) for row in A)