You are given an n x n
binary matrix grid
. You are allowed to change at most one 0
to be 1
.
Return the size of the largest island in grid
after applying this operation.
An island is a 4directionally connected group of 1
s.
Example 1:
Input: grid = [[1,0],[0,1]]
Output: 3
Explanation: Change one 0 to 1 and connect two 1s,
then we get an island with area = 3.
Example 2:
Input: grid = [[1,1],[1,0]]
Output: 4
Explanation: Change the 0 to 1 and make the island bigger,
only one island with area = 4.
Example 3:
Input: grid = [[1,1],[1,1]]
Output: 4
Explanation: Can't change any 0 to 1,
only one island with area = 4.
Constraints:
n == grid.length
n == grid[i].length
1 <= n <= 500
grid[i][j]
is either 0
or 1
.
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class Solution:
def largestIsland(self, grid: List[List[int]]) > int:
m, n = len(grid), len(grid[0])
# Label each island with an unique index
idx = 2
def dfs(i: int, j: int):
grid[i][j] = idx
size = 1
for dx, dy in [(0, 1), (0, 1), (1, 0), (1, 0)]:
x, y = i + dx, j + dy
if 0 <= x < m and 0 <= y < n:
if grid[x][y] == 1:
size += dfs(x, y)
return size
size_map = {} # island_idx: island_size
for i in range(m):
for j in range(n):
if grid[i][j] == 1:
island_size = dfs(i, j)
size_map[idx] = island_size
idx += 1
if not size_map:
# no island found
return 1
max_size = 0
for i in range(m):
for j in range(n):
if grid[i][j] == 0:
# get sizes of surrounding islands
islands = set()
for dx, dy in [(0, 1), (0, 1), (1, 0), (1, 0)]:
x, y = i + dx, j + dy
if 0 <= x < m and 0 <= y < n and grid[x][y] != 0:
islands.add(grid[x][y])
sizes = [size_map[island] for island in islands]
max_size = max(max_size, sum(sizes) + 1)
if max_size == 0:
# no water space
return m * n
return max_size
