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contain-virus.py
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contain-virus.py
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# Time: O((m * n)^(4/3)), days = O((m * n)^(1/3))
# Space: O(m * n)
# A virus is spreading rapidly, and your task is to quarantine the infected area by installing walls.
#
# The world is modeled as a 2-D array of cells,
# where 0 represents uninfected cells,
# and 1 represents cells contaminated with the virus.
# A wall (and only one wall) can be installed between any two 4-directionally adjacent cells,
# on the shared boundary.
#
# Every night, the virus spreads to all neighboring cells in all four directions unless blocked by a wall.
# Resources are limited.
# Each day, you can install walls around only one region --
# the affected area (continuous block of infected cells) that
# threatens the most uninfected cells the following night.
# There will never be a tie.
#
# Can you save the day? If so, what is the number of walls required?
# If not, and the world becomes fully infected, return the number of walls used.
#
# Example 1:
# Input: grid =
# [[0,1,0,0,0,0,0,1],
# [0,1,0,0,0,0,0,1],
# [0,0,0,0,0,0,0,1],
# [0,0,0,0,0,0,0,0]]
# Output: 10
# Explanation:
# There are 2 contaminated regions.
# On the first day, add 5 walls to quarantine the viral region on the left. The board after the virus spreads is:
#
# [[0,1,0,0,0,0,1,1],
# [0,1,0,0,0,0,1,1],
# [0,0,0,0,0,0,1,1],
# [0,0,0,0,0,0,0,1]]
#
# On the second day, add 5 walls to quarantine the viral region on the right. The virus is fully contained.
#
# Example 2:
# Input: grid =
# [[1,1,1],
# [1,0,1],
# [1,1,1]]
# Output: 4
# Explanation: Even though there is only one cell saved, there are 4 walls built.
# Notice that walls are only built on the shared boundary of two different cells.
#
# Example 3:
# Input: grid =
# [[1,1,1,0,0,0,0,0,0],
# [1,0,1,0,1,1,1,1,1],
# [1,1,1,0,0,0,0,0,0]]
# Output: 13
#
# Explanation: The region on the left only builds two new walls.
# Note:
# - The number of rows and columns of grid will each be in the range [1, 50].
# - Each grid[i][j] will be either 0 or 1.
# - Throughout the described process, there is always a contiguous viral region
# that will infect strictly more uncontaminated squares in the next round.
class Solution(object):
def containVirus(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
directions = [(0, 1), (0, -1), (-1, 0), (1, 0)]
def dfs(grid, r, c, lookup, regions, frontiers, perimeters):
if (r, c) in lookup:
return
lookup.add((r, c))
regions[-1].add((r, c))
for d in directions:
nr, nc = r+d[0], c+d[1]
if not (0 <= nr < len(grid) and \
0 <= nc < len(grid[r])):
continue
if grid[nr][nc] == 1:
dfs(grid, nr, nc, lookup, regions, frontiers, perimeters)
elif grid[nr][nc] == 0:
frontiers[-1].add((nr, nc))
perimeters[-1] += 1
result = 0
while True:
lookup, regions, frontiers, perimeters = set(), [], [], []
for r, row in enumerate(grid):
for c, val in enumerate(row):
if val == 1 and (r, c) not in lookup:
regions.append(set())
frontiers.append(set())
perimeters.append(0)
dfs(grid, r, c, lookup, regions, frontiers, perimeters)
if not regions: break
triage_idx = frontiers.index(max(frontiers, key = len))
for i, region in enumerate(regions):
if i == triage_idx:
result += perimeters[i]
for r, c in region:
grid[r][c] = -1
continue
for r, c in region:
for d in directions:
nr, nc = r+d[0], c+d[1]
if not (0 <= nr < len(grid) and \
0 <= nc < len(grid[r])):
continue
if grid[nr][nc] == 0:
grid[nr][nc] = 1
return result