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.
1var numIslands = function(grid) {
2 function dfs(grid, r, c) {
3 if (r < 0 || c < 0 || r >= grid.length || c >= grid[0].length || grid[r][c] === '0') {
4 return;
5 }
6 grid[r][c] = '0';
7 dfs(grid, r-1, c);
8 dfs(grid, r+1, c);
9 dfs(grid, r, c-1);
10 dfs(grid, r, c+1);
11 }
12 let numIslands = 0;
13 for (let r = 0; r < grid.length; r++) {
14 for (let c = 0; c < grid[0].length; c++) {
15 if (grid[r][c] === '1') {
16 numIslands++;
17 dfs(grid, r, c);
18 }
19 }
20 }
21 return numIslands;
22};The JavaScript solution leverages a classic DFS function dfs to explore the grid and perform in-place modification to mark visited cells. As it iterates through the grid, whenever it encounters '1', it initiates DFS to mark the whole island, counting it.
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.
1using System.Collections.Generic;
2
3public class Solution {
4 private static readonly int[][] directions = new int[][] {
5 new int[] {-1, 0}, new int[] {1, 0},
6 new int[] {0, -1}, new int[] {0, 1},
7 };
8
9 public int NumIslands(char[][] grid) {
10 if (grid == null || grid.Length == 0) {
11 return 0;
12 }
13 int numIslands = 0;
14 for (int r = 0; r < grid.Length; r++) {
15 for (int c = 0; c < grid[0].Length; c++) {
16 if (grid[r][c] == '1') {
17 numIslands++;
18 grid[r][c] = '0';
19 Queue<int[]> queue = new Queue<int[]>();
20 queue.Enqueue(new int[] {r, c});
21 while (queue.Count > 0) {
22 int[] cell = queue.Dequeue();
23 foreach (int[] d in directions) {
24 int nr = cell[0] + d[0], nc = cell[1] + d[1];
25 if (nr >= 0 && nr < grid.Length && nc >= 0 && nc < grid[0].Length && grid[nr][nc] == '1') {
26 queue.Enqueue(new int[] {nr, nc});
27 grid[nr][nc] = '0';
28 }
29 }
30 }
31 }
32 }
33 }
34 return numIslands;
35 }
36}C# employs a Queue to facilitate BFS exploration over the grid. For each '1' found, BFS iteratively processes nodes by marking discovered ones and queueing adjacent '1's to ensure the whole island is traversed.