Namaste, I'm Avinash Sharma.

Max Area of Island

published on 4/29/2022

You are given an m x n binary matrix grid. 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.

The area of an island is the number of cells with a value 1 in the island.

Return the maximum area of an island in grid. If there is no island, return 0.

Example 1

Input: grid = [[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
Explanation: The answer is not 11, because the island must be connected 4-directionally.

Example 2

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


  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 50
  • grid[i][j] is either 0 or 1.

Solution explanation:

  • First, we run dfs on the grid to find all the islands.
  • Then,for each island, we find the maximum area of each island.
  • And thne update the maxi with the maximum area for the island.
  • Finally, we return the maxi.
class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
n,m,maxi = len(grid),len(grid[0]),0
def dfs(i, j):
if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] != 1:
return 0
maxArea = 1
grid[i][j] = '#' # this will act as visited set
maxArea += dfs(i+1, j)
maxArea += dfs(i-1, j)
maxArea += dfs(i, j+1)
maxArea += dfs(i, j-1)
return maxArea
for i in range(n):
for j in range(m):
if grid[i][j] == 1:
maxi = max(dfs(i,j),maxi)
return maxi
Created with ❤️ by Avinash Sharma