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🟨 Number of Islands (#200)

🔗 Problem Link​

📋 Problem Statement​

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

💡 Examples​

Example 1​

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2​

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

🔑 Key Insights & Approach​

Core Observation: Use DFS or BFS to mark all connected land cells as visited. Each DFS/BFS call represents one island.

Why DFS/BFS?

  • O(m * n) time
  • O(min(m, n)) space for BFS, O(m * n) for DFS worst case
  • Simple and efficient

Approaches:

  1. DFS: O(m * n) time, O(m * n) space
  2. BFS: O(m * n) time, O(min(m, n)) space
  3. Union-Find: O(m * n) time with path compression

Pattern: "Graph Connected Components" using DFS/BFS.

🐍 Solution: Python​

Approach 1: DFS​

Time Complexity: O(m * n) | Space Complexity: O(m * n)

from typing import List

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        if not grid:
            return 0

        rows, cols = len(grid), len(grid[0])
        islands = 0

        def dfs(r, c):
            if (r < 0 or r >= rows or c < 0 or c >= cols or
                grid[r][c] == '0'):
                return

            grid[r][c] = '0'  # Mark as visited

            # Explore 4 directions
            dfs(r + 1, c)
            dfs(r - 1, c)
            dfs(r, c + 1)
            dfs(r, c - 1)

        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == '1':
                    islands += 1
                    dfs(r, c)

        return islands

Approach 2: BFS​

Time Complexity: O(m * n) | Space Complexity: O(min(m, n))

from collections import deque
from typing import List

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        if not grid:
            return 0

        rows, cols = len(grid), len(grid[0])
        islands = 0

        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == '1':
                    islands += 1

                    queue = deque([(r, c)])
                    grid[r][c] = '0'

                    while queue:
                        row, col = queue.popleft()

                        for dr, dc in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                            nr, nc = row + dr, col + dc
                            if (0 <= nr < rows and 0 <= nc < cols and
                                grid[nr][nc] == '1'):
                                queue.append((nr, nc))
                                grid[nr][nc] = '0'

        return islands

🔵 Solution: Golang​

Approach: DFS​

Time Complexity: O(m * n) | Space Complexity: O(m * n)

func numIslands(grid [][]byte) int {
    if len(grid) == 0 {
        return 0
    }

    rows, cols := len(grid), len(grid[0])
    islands := 0

    var dfs func(r, c int)
    dfs = func(r, c int) {
        if r < 0 || r >= rows || c < 0 || c >= cols || grid[r][c] == '0' {
            return
        }

        grid[r][c] = '0'

        dfs(r+1, c)
        dfs(r-1, c)
        dfs(r, c+1)
        dfs(r, c-1)
    }

    for r := 0; r < rows; r++ {
        for c := 0; c < cols; c++ {
            if grid[r][c] == '1' {
                islands++
                dfs(r, c)
            }
        }
    }

    return islands
}

📚 References​