This approach uses a depth-first search (DFS) to mark land cells connected to the boundaries. We iterate over the boundary cells of the grid and if a land cell is found, we perform DFS to mark all connected land cells as visited. Finally, we count all remaining unvisited land cells inside the grid, which are considered enclaves.
Time Complexity: O(m * n) as each cell is visited at most once.
Space Complexity: O(m * n) in the worst case due to recursion stack used by DFS.
1class Solution:
2 def numEnclaves(self, grid: List[List[int]]) -> int:
3 def dfs(i: int, j: int):
4 if i < 0 or j < 0 or i >= len(grid) or j >= len(grid[0]) or grid[i][j] == 0:
5 return
6 grid[i][j] = 0
7 dfs(i + 1, j)
8 dfs(i - 1, j)
9 dfs(i, j + 1)
10 dfs(i, j - 1)
11
12 m, n = len(grid), len(grid[0])
13 for i in range(m):
14 if grid[i][0] == 1:
15 dfs(i, 0)
16 if grid[i][n - 1] == 1:
17 dfs(i, n - 1)
18 for j in range(n):
19 if grid[0][j] == 1:
20 dfs(0, j)
21 if grid[m - 1][j] == 1:
22 dfs(m - 1, j)
23
24 return sum(grid[i][j] for i in range(1, m - 1) for j in range(1, n - 1) if grid[i][j] == 1)
25In this Python implementation, DFS is defined as an inner function. The method iterates over boundary cells, marking connected land cells. Finally, it sums remaining inner land cells (enclaves).
In this approach, we use a queue to perform BFS to mark land cells connected to the boundaries. By enqueueing each boundary land cell and exploring its neighbors iteratively, we can mark all reachable boundary-connected lands. This is followed by counting the unvisited land cells – those are the enclaves.
Time Complexity: O(m * n).
Space Complexity: O(min(m, n)) considering the queue size in the worst case.
1var numEnclaves = function(grid) {
2 const bfs = (startRow, startCol) => {
3 const queue = [];
4 queue.push([startRow, startCol]);
5 grid[startRow][startCol] = 0;
6 const directions = [[1, 0], [-1, 0], [0, 1], [0, -1]];
7
8 while (queue.length !== 0) {
9 const [r, c] = queue.shift();
10 for (const [dx, dy] of directions) {
11 const newRow = r + dx;
12 const newCol = c + dy;
13 if (newRow >= 0 && newRow < grid.length && newCol >= 0 && newCol < grid[0].length && grid[newRow][newCol] === 1) {
14 grid[newRow][newCol] = 0;
15 queue.push([newRow, newCol]);
16 }
17 }
18 }
19 };
20
21 const m = grid.length, n = grid[0].length;
22
23 for (let i = 0; i < m; i++) {
24 if (grid[i][0] === 1) bfs(i, 0);
25 if (grid[i][n - 1] === 1) bfs(i, n - 1);
26 }
27 for (let j = 0; j < n; j++) {
28 if (grid[0][j] === 1) bfs(0, j);
29 if (grid[m - 1][j] === 1) bfs(m - 1, j);
30 }
31
32 let count = 0;
33 for (let i = 0; i < m; i++) {
34 for (let j = 0; j < n; j++) {
35 if (grid[i][j] === 1) {
36 count++;
37 }
38 }
39 }
40 return count;
41};
42This JavaScript solution performs BFS starting from boundary islands using a queue. The queue helps in exploring nodes layer by layer (BFS style), marking reachable nodes. Remaining unmarked are counted as enclaves.