This approach uses Depth-First Search (DFS) to explore the grid. We iterate over each cell in the grid, and every time we find an unvisited '1', it indicates the discovery of a new island. We then perform DFS from that cell to mark all the connected '1's as visited, effectively marking the entire island.
Time Complexity: O(m*n) where m is the number of rows and n is the number of columns since each cell is visited once.
Space Complexity: O(m*n) in the worst case due to the recursion stack used by DFS.
1class Solution:
2 def dfs(self, grid, r, c):
3 if r < 0 or c < 0 or r >= len(grid) or c >= len(grid[0]) or grid[r][c] == '0':
4 return
5 grid[r][c] = '0'
6 self.dfs(grid, r-1, c)
7 self.dfs(grid, r+1, c)
8 self.dfs(grid, r, c-1)
9 self.dfs(grid, r, c+1)
10
11 def numIslands(self, grid):
12 if not grid:
13 return 0
14 num_islands = 0
15 for r in range(len(grid)):
16 for c in range(len(grid[0])):
17 if grid[r][c] == '1':
18 num_islands += 1
19 self.dfs(grid, r, c)
20 return num_islandsPython's solution uses a method dfs to explore and mark connected lands. We iterate over the entire grid, and for every '1' encountered, initiate a DFS from that position, effectively exploring the whole island.
This approach uses Breadth-First Search (BFS) to traverse the grid. Similar to DFS, we treat each '1' as a node in a graph. On encountering a '1', we initiate a BFS to explore all connected '1's (island nodes) by utilizing a queue for the frontier, marking them in-place as visited.
Time Complexity: O(m*n), where m and n are rows and columns.#
Space Complexity: O(min(m, n)) due to the queue storage in BFS.
1import java.util.LinkedList;
2import java.util.Queue;
3
4public class Solution {
5 private static final int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
6
7 public int numIslands(char[][] grid) {
8 if (grid == null || grid.length == 0) {
9 return 0;
10 }
11 int numIslands = 0;
12 for (int r = 0; r < grid.length; r++) {
13 for (int c = 0; c < grid[0].length; c++) {
14 if (grid[r][c] == '1') {
15 numIslands++;
16 grid[r][c] = '0';
17 Queue<int[]> queue = new LinkedList<>();
18 queue.add(new int[]{r, c});
19 while (!queue.isEmpty()) {
20 int[] cell = queue.poll();
21 for (int[] d : directions) {
22 int nr = cell[0] + d[0], nc = cell[1] + d[1];
23 if (nr >= 0 && nr < grid.length && nc >= 0 && nc < grid[0].length && grid[nr][nc] == '1') {
24 queue.add(new int[]{nr, nc});
25 grid[nr][nc] = '0';
26 }
27 }
28 }
29 }
30 }
31 }
32 return numIslands;
33 }
34}Java's solution leverages a Queue to implement BFS. When encountering '1', BFS starts and expands, visiting the whole component marking each cell and exploring adjacent '1's iteratively using the queue.