The challenge
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 ``.
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
Constraints:
m == grid.length
n == grid[i].length
1 <= m, n <= 50
grid[i][j]
is either `` or1
.
The solution in Python code
class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
def get_neighbor(pos,grid):
y,x = pos
ns = []
if x>=1:
ns.append((y,x-1))
if x<len(grid[0])-1:
ns.append((y,x+1))
if y>=1:
ns.append((y-1,x))
if y<len(grid)-1:
ns.append((y+1,x))
return ns
marked = set()
land = []
for row in range(len(grid)):
for col in range(len(grid[row])):
if grid[row][col] == 1 and (row,col) not in marked:
curr_land_len = 1
marked.add((row,col))
stack = [(row,col)]
while stack:
current = stack.pop()
neighbor = get_neighbor(current,grid)
for n in neighbor:
y,x = n
if grid[y][x] == 1 and (y,x) not in marked:
marked.add((y,x))
curr_land_len += 1
stack.append((y,x))
land.append(curr_land_len)
return (max(land) if len(land)!=0 else 0)