Max Area of Island

DFS

Problem

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Note: The length of each dimension in the given grid does not exceed 50.

For example:

Input:
[[0,0,1,0,0,0,0,1,0,0,0,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,1,1,0,1,0,0,0,0,0,0,0,0],
 [0,1,0,0,1,1,0,0,1,0,1,0,0],
 [0,1,0,0,1,1,0,0,1,1,1,0,0],
 [0,0,0,0,0,0,0,0,0,0,1,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,0,0,0,0,0,0,1,1,0,0,0,0]]
 
 Output: 6 
 Note the answer is not 11, because the 
 island must be connected 4-directionally.

Input: [[0,0,0,0,0,0,0,0]]
Output: 0

Thought Process

• Apply DFS, similar to Number of Islands, since we are dealing with connected components

• Since we are dealing with area, we will just count the number of 1's we come across

Solution

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        
        maxArea = float("-inf")
        
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                self.count = 0
                if grid[i][j] == 1:
                    area = self.dfsearch(grid, i, j)
                    maxArea = max(maxArea, area)
                    
        return maxArea if maxArea != float("-inf") else 0
    
    def dfsearch(self, grid, i, j):
        if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] != 1:
            return 0
        
        grid[i][j] = 0
        return 1 + self.dfsearch(grid, i+1, j) + self.dfsearch(grid, i-1, j) + self.dfsearch(grid, i, j-1) + self.dfsearch(grid, i, j+1)
        
        
        

Time Complexity

• Time: O(m*n)

• Space: O(m*n) for the recursion call stack cause worst case is we have all 1s