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.
1#include <stdbool.h>
2
3void dfs(char** grid, int r, int c, int gridSize, int* gridColSize) {
4 if (r < 0 || c < 0 || r >= gridSize || c >= gridColSize[r] || grid[r][c] == '0') {
5 return;
6 }
7 grid[r][c] = '0';
8 dfs(grid, r-1, c, gridSize, gridColSize);
9 dfs(grid, r+1, c, gridSize, gridColSize);
10 dfs(grid, r, c-1, gridSize, gridColSize);
11 dfs(grid, r, c+1, gridSize, gridColSize);
12}
13
14int numIslands(char** grid, int gridSize, int* gridColSize) {
15 int numIslands = 0;
16 for (int i = 0; i < gridSize; i++) {
17 for (int j = 0; j < gridColSize[i]; j++) {
18 if (grid[i][j] == '1') {
19 numIslands++;
20 dfs(grid, i, j, gridSize, gridColSize);
21 }
22 }
23 }
24 return numIslands;
25}The function dfs is responsible for marking connected '1's as visited by changing them to '0'. The numIslands function traverses each cell in the grid, and upon finding a '1', it increments the count and triggers a DFS from that cell to mark the entire 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.
1from collections import deque
2
3class Solution:
4 def numIslands(self, grid):
5 if not grid:
6 return 0
7 num_islands = 0
8 rows, cols = len(grid), len(grid[0])
9
10 def bfs(r, c):
11 queue = deque([(r, c)])
12 grid[r][c] = '0'
13 directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
14 while queue:
15 cr, cc = queue.popleft()
16 for dr, dc in directions:
17 nr, nc = cr + dr, cc + dc
18 if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] == '1':
19 grid[nr][nc] = '0'
20 queue.append((nr, nc))
21
22 for r in range(rows):
23 for c in range(cols):
24 if grid[r][c] == '1':
25 num_islands += 1
26 bfs(r, c)
27 return num_islandsThe Python implementation employs the collections.deque to perform BFS. Upon encountering a '1', the coordinates are queued and BFS enqueues its adjacent lands, effectively exploring the entire island and marking progress.