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0200-number-of-islands.py
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0200-number-of-islands.py
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class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid or not grid[0]:
return 0
islands = 0
visit = set()
rows, cols = len(grid), len(grid[0])
def dfs(r, c):
if (
r not in range(rows)
or c not in range(cols)
or grid[r][c] == "0"
or (r, c) in visit
):
return
visit.add((r, c))
directions = [[0, 1], [0, -1], [1, 0], [-1, 0]]
for dr, dc in directions:
dfs(r + dr, c + dc)
for r in range(rows):
for c in range(cols):
if grid[r][c] == "1" and (r, c) not in visit:
islands += 1
dfs(r, c)
return islands
# BFS Version From Video
class SolutionBFS:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid:
return 0
rows, cols = len(grid), len(grid[0])
visited=set()
islands=0
def bfs(r,c):
q = deque()
visited.add((r,c))
q.append((r,c))
while q:
row,col = q.popleft()
directions= [[1,0],[-1,0],[0,1],[0,-1]]
for dr,dc in directions:
r,c = row + dr, col + dc
if (r) in range(rows) and (c) in range(cols) and grid[r][c] == '1' and (r ,c) not in visited:
q.append((r , c ))
visited.add((r, c ))
for r in range(rows):
for c in range(cols):
if grid[r][c] == "1" and (r,c) not in visited:
bfs(r,c)
islands +=1
return islands