Max Area of Island

Given a 2D array grid, where zeroes represent water and ones represent land, return the size of the largest island.

Note: An island is one or more cells in grid containing a value of one that may be connected vertically or horizontally. Each cell in an island contributes a value of one to the current island's size.

Example(s)

Example 1:

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

Output: 4
Explanation:
There are two islands:
- Island 1: cells (0,0), (0,1), (1,0), (1,1) → size 4
- Island 2: cell (2,2) → size 1
The largest island has size 4.

Example 2:

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

Output: 4
Explanation:
The largest island has 4 connected cells.

Example 3:

Input: grid = [
  [1, 1, 1],
  [1, 1, 1],
  [1, 1, 1],
]

Output: 9
Explanation:
All cells form one large island of size 9.

Example 4:

Input: grid = [
  [0, 0, 0],
  [0, 0, 0],
  [0, 0, 0],
]

Output: 0
Explanation:
No islands (all water).

Solution

The solution uses DFS (Depth-First Search) or BFS (Breadth-First Search) to explore each island:

Iterate through grid: For each cell that is land (1) and not visited
Explore island: Use DFS/BFS to visit all connected land cells
Count size: Count how many cells are in the current island
Track maximum: Keep track of the largest island size seen so far
Mark visited: Mark cells as visited to avoid counting them twice

JavaScript Solution
Python Solution

JavaScript Solution - DFS

/**
* Find the size of the largest island
* @param {number[][]} grid - 2D array where 1 = land, 0 = water
* @return {number} - Size of the largest island
*/
function maxAreaOfIsland(grid) {
if (!grid || grid.length === 0) return 0;

const rows = grid.length;
const cols = grid[0].length;
let maxArea = 0;

/**
 * DFS to explore an island and count its size
 * @param {number} row - Current row
 * @param {number} col - Current column
 * @return {number} - Size of the island
 */
function dfs(row, col) {
  // Base cases: out of bounds or water or already visited
  if (
    row < 0 || row >= rows ||
    col < 0 || col >= cols ||
    grid[row][col] === 0
  ) {
    return 0;
  }
  
  // Mark current cell as visited (set to 0)
  grid[row][col] = 0;
  
  // Count current cell and explore neighbors
  let area = 1;
  area += dfs(row - 1, col); // up
  area += dfs(row + 1, col); // down
  area += dfs(row, col - 1); // left
  area += dfs(row, col + 1); // right
  
  return area;
}

// Iterate through all cells
for (let i = 0; i < rows; i++) {
  for (let j = 0; j < cols; j++) {
    if (grid[i][j] === 1) {
      // Found a new island, explore it
      const area = dfs(i, j);
      maxArea = Math.max(maxArea, area);
    }
  }
}

return maxArea;
}

// Test cases
const grid1 = [
[1, 1, 0],
[1, 1, 0],
[0, 0, 1],
];
console.log('Example 1:', maxAreaOfIsland(grid1)); // 4

const grid2 = [
[0, 0, 1, 0, 0],
[0, 0, 0, 1, 1],
[0, 1, 1, 0, 0],
[0, 1, 0, 0, 1],
[0, 1, 0, 0, 1],
];
console.log('Example 2:', maxAreaOfIsland(grid2)); // 4

const grid3 = [
[1, 1, 1],
[1, 1, 1],
[1, 1, 1],
];
console.log('Example 3:', maxAreaOfIsland(grid3)); // 9

const grid4 = [
[0, 0, 0],
[0, 0, 0],
[0, 0, 0],
];
console.log('Example 4:', maxAreaOfIsland(grid4)); // 0

Output:

Click "Run Code" to execute the code and see the results.

Python Solution - DFS

from typing import List

def max_area_of_island(grid: List[List[int]]) -> int:
  """
  Find the size of the largest island
  
  Args:
      grid: 2D array where 1 = land, 0 = water
      
  Returns:
      int: Size of the largest island
  """
  if not grid or not grid[0]:
      return 0
  
  rows = len(grid)
  cols = len(grid[0])
  max_area = 0
  
  def dfs(row: int, col: int) -> int:
      """
      DFS to explore an island and count its size
      
      Returns:
          int: Size of the island
      """
      # Base cases: out of bounds or water or already visited
      if (
          row < 0 or row >= rows or
          col < 0 or col >= cols or
          grid[row][col] == 0
      ):
          return 0
      
      # Mark current cell as visited (set to 0)
      grid[row][col] = 0
      
      # Count current cell and explore neighbors
      area = 1
      area += dfs(row - 1, col)  # up
      area += dfs(row + 1, col)  # down
      area += dfs(row, col - 1)  # left
      area += dfs(row, col + 1)  # right
      
      return area
  
  # Iterate through all cells
  for i in range(rows):
      for j in range(cols):
          if grid[i][j] == 1:
              # Found a new island, explore it
              area = dfs(i, j)
              max_area = max(max_area, area)
  
  return max_area

# Test cases
grid1 = [
  [1, 1, 0],
  [1, 1, 0],
  [0, 0, 1],
]
print('Example 1:', max_area_of_island(grid1))  # 4

grid2 = [
  [0, 0, 1, 0, 0],
  [0, 0, 0, 1, 1],
  [0, 1, 1, 0, 0],
  [0, 1, 0, 0, 1],
  [0, 1, 0, 0, 1],
]
print('Example 2:', max_area_of_island(grid2))  # 4

grid3 = [
  [1, 1, 1],
  [1, 1, 1],
  [1, 1, 1],
]
print('Example 3:', max_area_of_island(grid3))  # 9

grid4 = [
  [0, 0, 0],
  [0, 0, 0],
  [0, 0, 0],
]
print('Example 4:', max_area_of_island(grid4))  # 0

Loading Python runtime...

Output:

Click "Run Code" to execute the code and see the results.

Alternative Solution (BFS)

Here's a BFS (Breadth-First Search) approach:

JavaScript BFS
Python BFS

/**
 * BFS approach
 */
function maxAreaOfIslandBFS(grid) {
  if (!grid || grid.length === 0) return 0;
  
  const rows = grid.length;
  const cols = grid[0].length;
  let maxArea = 0;
  const directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]; // up, down, left, right
  
  function bfs(startRow, startCol) {
    const queue = [[startRow, startCol]];
    grid[startRow][startCol] = 0; // Mark as visited
    let area = 0;
    
    while (queue.length > 0) {
      const [row, col] = queue.shift();
      area++;
      
      // Explore neighbors
      for (const [dr, dc] of directions) {
        const newRow = row + dr;
        const newCol = col + dc;
        
        if (
          newRow >= 0 && newRow < rows &&
          newCol >= 0 && newCol < cols &&
          grid[newRow][newCol] === 1
        ) {
          grid[newRow][newCol] = 0; // Mark as visited
          queue.push([newRow, newCol]);
        }
      }
    }
    
    return area;
  }
  
  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      if (grid[i][j] === 1) {
        const area = bfs(i, j);
        maxArea = Math.max(maxArea, area);
      }
    }
  }
  
  return maxArea;
}

from collections import deque

def max_area_of_island_bfs(grid: List[List[int]]) -> int:
    """
    BFS approach
    """
    if not grid or not grid[0]:
        return 0
    
    rows = len(grid)
    cols = len(grid[0])
    max_area = 0
    directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]  # up, down, left, right
    
    def bfs(start_row: int, start_col: int) -> int:
        queue = deque([(start_row, start_col)])
        grid[start_row][start_col] = 0  # Mark as visited
        area = 0
        
        while queue:
            row, col = queue.popleft()
            area += 1
            
            # Explore neighbors
            for dr, dc in directions:
                new_row = row + dr
                new_col = col + dc
                
                if (
                    0 <= new_row < rows and
                    0 <= new_col < cols and
                    grid[new_row][new_col] == 1
                ):
                    grid[new_row][new_col] = 0  # Mark as visited
                    queue.append((new_row, new_col))
        
        return area
    
    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 1:
                area = bfs(i, j)
                max_area = max(max_area, area)
    
    return max_area

Alternative: Without Modifying Original Grid

If we need to preserve the original grid, we can use a visited set:

JavaScript with Visited Set
Python with Visited Set

/**
 * Solution that doesn't modify the original grid
 */
function maxAreaOfIslandPreserve(grid) {
  if (!grid || grid.length === 0) return 0;
  
  const rows = grid.length;
  const cols = grid[0].length;
  const visited = new Set();
  let maxArea = 0;
  
  function dfs(row, col) {
    const key = `${row},${col}`;
    
    if (
      row < 0 || row >= rows ||
      col < 0 || col >= cols ||
      grid[row][col] === 0 ||
      visited.has(key)
    ) {
      return 0;
    }
    
    visited.add(key);
    let area = 1;
    area += dfs(row - 1, col);
    area += dfs(row + 1, col);
    area += dfs(row, col - 1);
    area += dfs(row, col + 1);
    
    return area;
  }
  
  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      if (grid[i][j] === 1 && !visited.has(`${i},${j}`)) {
        const area = dfs(i, j);
        maxArea = Math.max(maxArea, area);
      }
    }
  }
  
  return maxArea;
}

def max_area_of_island_preserve(grid: List[List[int]]) -> int:
    """
    Solution that doesn't modify the original grid
    """
    if not grid or not grid[0]:
        return 0
    
    rows = len(grid)
    cols = len(grid[0])
    visited = set()
    max_area = 0
    
    def dfs(row: int, col: int) -> int:
        if (
            row < 0 or row >= rows or
            col < 0 or col >= cols or
            grid[row][col] == 0 or
            (row, col) in visited
        ):
            return 0
        
        visited.add((row, col))
        area = 1
        area += dfs(row - 1, col)
        area += dfs(row + 1, col)
        area += dfs(row, col - 1)
        area += dfs(row, col + 1)
        
        return area
    
    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 1 and (i, j) not in visited:
                area = dfs(i, j)
                max_area = max(max_area, area)
    
    return max_area

Complexity

Time Complexity: O(m × n) - Where m and n are the dimensions of the grid. We visit each cell at most once.
Space Complexity: O(m × n) - For DFS recursion stack in worst case (when entire grid is one island). For BFS, it's the queue size. If we use a visited set, it's also O(m × n).

Approach

The solution uses graph traversal (DFS or BFS):

Iterate through grid: For each cell (i, j)
Check if land: If grid[i][j] === 1 and not visited
Explore island: Use DFS/BFS to visit all connected land cells
Count cells: Count how many cells are in the current island
Update maximum: Track the largest island size
Mark visited: Mark cells as visited (set to 0 or use visited set)

Key Insights

Connected components: Islands are connected components in a graph
4-directional: Cells are connected horizontally and vertically (not diagonally)
DFS vs BFS: Both work, DFS is simpler to implement recursively
In-place marking: Setting visited cells to 0 saves space
Single pass: Each cell is visited at most once

Step-by-Step Example

Let's trace through Example 1: grid = [[1,1,0],[1,1,0],[0,0,1]]

Initial grid:
[1, 1, 0]
[1, 1, 0]
[0, 0, 1]

Step 1: Found land at (0,0)
  DFS from (0,0):
    - Visit (0,0) → area = 1
    - Visit (0,1) → area = 2
    - Visit (1,0) → area = 3
    - Visit (1,1) → area = 4
    - No more neighbors
  Island size = 4
  maxArea = 4

Step 2: Found land at (2,2)
  DFS from (2,2):
    - Visit (2,2) → area = 1
    - No neighbors
  Island size = 1
  maxArea = max(4, 1) = 4

Final: maxArea = 4

Visual Representation

Grid:           DFS exploration:
[1, 1, 0]       [✓, ✓, 0]
[1, 1, 0]  →    [✓, ✓, 0]
[0, 0, 1]       [0, 0, ✓]

Island 1: 4 cells (marked with ✓)
Island 2: 1 cell (marked with ✓)
Largest: 4

Edge Cases

Empty grid: Return 0
All water: Return 0
All land: Return m × n (entire grid is one island)
Single cell land: Return 1
Single cell water: Return 0
No islands: Return 0

Optimization Notes

Early termination: Not applicable (need to check all islands)
In-place marking: More space-efficient than visited set
DFS vs BFS: DFS is typically preferred for this problem (simpler code)
Direction array: Can use [[-1,0],[1,0],[0,-1],[0,1]] for cleaner code

Number of Islands: Count total number of islands
Island Perimeter: Calculate perimeter of islands
Surrounded Regions: Mark regions surrounded by water
Walls and Gates: BFS from multiple sources
Flood Fill: Similar DFS/BFS pattern

Takeaways

DFS/BFS are essential for grid traversal problems
Connected components can be found using graph traversal
In-place marking saves space (set to 0 instead of visited set)
4-directional means up, down, left, right (not diagonals)
Single pass through grid is optimal

Example(s)​

Solution​

JavaScript Solution - DFS

Output:

Python Solution - DFS

Output:

Alternative Solution (BFS)​

Alternative: Without Modifying Original Grid​

Complexity​

Approach​

Key Insights​

Step-by-Step Example​

Visual Representation​

Edge Cases​

Optimization Notes​

Related Problems​

Takeaways​

Example(s)

Solution

Alternative Solution (BFS)

Alternative: Without Modifying Original Grid

Complexity

Approach

Key Insights

Step-by-Step Example

Visual Representation

Edge Cases

Optimization Notes

Related Problems

Takeaways