Number of Islands

Description:​

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.

Video Explanation:​

Approaches:​

1. Depth-First Search (DFS) (Optimal)​

This problem essentially asks us to find the number of Connected Components in an undirected graph disguised as a 2D matrix. Every '1' represents a land node, and it is connected to adjacent '1's.

We can solve this by iterating through every cell in the grid. Whenever we encounter a '1', it means we have found a new, unvisited island. We increment our island count, and then use a Depth-First Search (DFS) to traverse the entire island, marking all connected '1's as visited so we don't double-count them later.

Algorithm:

1. Initialize an island_count to 0.
2. Iterate through every cell (r, c) in the m x n grid.
3. If the current cell is '1' (land):
   • Increment island_count by 1.
   • Launch a dfs(r, c) helper function to explore the island.
4. In the dfs function:
   • Base Case: If the current row r or column c is out of bounds, or if the current cell is '0' (water), return immediately.
   • Mark the current cell as visited by changing its value from '1' to '0'. This modifies the grid in-place to save memory (alternatively, you could use a separate visited boolean matrix).
   • Recursively call dfs on all 4 adjacent directions (up, down, left, right).
5. Return the final island_count.

Complexity​

• Time Complexity: $O(M \times N)$ where $M$ is the number of rows and $N$ is the number of columns. We visit every cell in the grid a constant number of times.
• Space Complexity: $O(M \times N)$ in the worst case for the recursion call stack. This happens if the entire grid is filled with '1's, meaning the DFS goes $M \times N$ levels deep.

Solutions:​

C++

class Solution {
private:
    void dfs(vector<vector<char>>& grid, int r, int c) {
        // Boundary checks and water check
        if (r < 0 || c < 0 || r >= grid.size() || c >= grid[0].size() || grid[r][c] == '0') {
            return;
        }
        
        // Mark as visited by sinking the land
        grid[r][c] = '0';
        
        // Traverse all 4 adjacent directions
        dfs(grid, r - 1, c); // Up
        dfs(grid, r + 1, c); // Down
        dfs(grid, r, c - 1); // Left
        dfs(grid, r, c + 1); // Right
    }
    
public:
    int numIslands(vector<vector<char>>& grid) {
        if (grid.empty()) return 0;
        
        int count = 0;
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j] == '1') {
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        
        return count;
    }
};

Java

class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) return 0;
        
        int count = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') {
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        
        return count;
    }
    
    private void dfs(char[][] grid, int r, int c) {
        // Boundary checks and water check
        if (r < 0 || c < 0 || r >= grid.length || c >= grid[0].length || grid[r][c] == '0') {
            return;
        }
        
        // Mark as visited by sinking the land
        grid[r][c] = '0';
        
        // Traverse all 4 adjacent directions
        dfs(grid, r - 1, c); // Up
        dfs(grid, r + 1, c); // Down
        dfs(grid, r, c - 1); // Left
        dfs(grid, r, c + 1); // Right
    }
}

Python

class Solution:
    def numIslands(self, grid: list[list[str]]) -> int:
        if not grid:
            return 0
            
        count = 0
        rows, cols = len(grid), len(grid[0])
        
        def dfs(r, c):
            # Boundary checks and water check
            if r < 0 or c < 0 or r >= rows or c >= cols or grid[r][c] == '0':
                return
                
            # Mark as visited by sinking the land
            grid[r][c] = '0'
            
            # Traverse all 4 adjacent directions
            dfs(r - 1, c) # Up
            dfs(r + 1, c) # Down
            dfs(r, c - 1) # Left
            dfs(r, c + 1) # Right
            
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == '1':
                    count += 1
                    dfs(r, c)
                    
        return count

JavaScript

/**
 * @param {character[][]} grid
 * @return {number}
 */
const numIslands = function(grid) {
    if (!grid || grid.length === 0) return 0;
    
    let count = 0;
    const rows = grid.length;
    const cols = grid[0].length;
    
    const dfs = (r, c) => {
        // Boundary checks and water check
        if (r < 0 || c < 0 || r >= rows || c >= cols || grid[r][c] === '0') {
            return;
        }
        
        // Mark as visited by sinking the land
        grid[r][c] = '0';
        
        // Traverse all 4 adjacent directions
        dfs(r - 1, c); // Up
        dfs(r + 1, c); // Down
        dfs(r, c - 1); // Left
        dfs(r, c + 1); // Right
    };
    
    for (let r = 0; r < rows; r++) {
        for (let c = 0; c < cols; c++) {
            if (grid[r][c] === '1') {
                count++;
                dfs(r, c);
            }
        }
    }
    
    return count;
};