Making a large island¶
Time: O(N^2); Space: O(N^2); hard
In a 2D grid of 0s and 1s, we change at most one 0 to a 1.
After, what is the size of the largest island? (An island is a 4-directionally connected group of 1s).
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:
1 <= len(grid) = len(grid[0]) <= 50
0 <= grid[i][j] <= 1
[1]:
class Solution1(object):
"""
Time: O(N^2)
Space: O(N^2)
"""
def largestIsland(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
directions = [(0, -1), (0, 1), (-1, 0), (1, 0)]
def dfs(r, c, index, grid):
if not (0 <= r < len(grid) and
0 <= c < len(grid[0]) and
grid[r][c] == 1):
return 0
result = 1
grid[r][c] = index
for d in directions:
result += dfs(r+d[0], c+d[1], index, grid)
return result
area = {}
index = 2
for r in range(len(grid)):
for c in range(len(grid[r])):
if grid[r][c] == 1:
area[index] = dfs(r, c, index, grid)
index += 1
result = max(area.values() or [0])
for r in range(len(grid)):
for c in range(len(grid[r])):
if grid[r][c] == 0:
seen = set()
for d in directions:
nr, nc = r+d[0], c+d[1]
if not (0 <= nr < len(grid) and
0 <= nc < len(grid[0]) and
grid[nr][nc] > 1):
continue
seen.add(grid[nr][nc])
result = max(result, 1 + sum(area[i] for i in seen))
return result
[2]:
s = Solution1()
grid = [[1, 0], [0, 1]]
assert s.largestIsland(grid) == 3
grid = [[1, 1], [1, 0]]
assert s.largestIsland(grid) == 4
grid = [[1, 1], [1, 1]]
assert s.largestIsland(grid) == 4