Hard
Number of Distinct Islands II — Python
Full explanation · Time O((m * n) * log(m * n)) · Space O(m * n)
# Time: O((m * n) * log(m * n))
# Space: O(m * n)
class Solution(object):
def numDistinctIslands2(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
directions = [(0, -1), (0, 1), (-1, 0), (1, 0)]
def dfs(i, j, grid, island):
if not (0 <= i < len(grid) and \
0 <= j < len(grid[0]) and \
grid[i][j] > 0):
return False
grid[i][j] *= -1
island.append((i, j))
for d in directions:
dfs(i+d[0], j+d[1], grid, island)
return True
def normalize(island):
shapes = [[] for _ in xrange(8)]
for x, y in island:
rotations_and_reflections = [[ x, y], [ x, -y], [-x, y], [-x, -y],
[ y, x], [ y, -x], [-y, x], [-y, -x]]
for i in xrange(len(rotations_and_reflections)):
shapes[i].append(rotations_and_reflections[i])
for shape in shapes:
shape.sort() # Time: O(ilogi), i is the size of the island, the max would be (m * n)
origin = list(shape[0])
for p in shape:
p[0] -= origin[0]
p[1] -= origin[1]
return min(shapes)
islands = set()
for i in xrange(len(grid)):
for j in xrange(len(grid[0])):
island = []
if dfs(i, j, grid, island):
islands.add(str(normalize(island)))
return len(islands)