[
bfs
dfs
graph
]
BinarySearch 0048 Sinking Islands
Problem statement
https://binarysearch.com/problems/Sinking-Islands/
Solution
The idea is to add all border with values 1
to queue and pefrome bfs (or stack with dfs) and mark all visited nodes as 2
. Then run once again, where we substitute all 1
with 0
, our sinked islands and finally replace all 2
with 1
.
Complexity
It is O(mn)
for time and space.
Code
class Solution:
def solve(self, grid):
n, m = len(grid[0]), len(grid)
b1 = [(0, i) for i in range(n)] + [(i, 0) for i in range(1, m)]
b2 = [(m-1, i) for i in range(n)] + [(i, n-1) for i in range(m-1)]
V = set()
q = deque([(x, y) for x, y in b1 + b2 if grid[x][y] == 1])
while q:
i, j = q.popleft()
grid[i][j] = 2
for x, y in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if not 0 <= x < m or not 0 <= y < n or grid[x][y] != 1 or (x, y) in V: continue
V.add((x, y))
q += [(x, y)]
for i, j in product(range(m), range(n)):
if grid[i][j] == 1: grid[i][j] = 0
for i, j in product(range(m), range(n)):
if grid[i][j] == 2: grid[i][j] = 1
return grid