Leetcode #749: Contain Virus
In this guide, we solve Leetcode #749 Contain Virus in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A virus is spreading rapidly, and your task is to quarantine the infected area by installing walls. The world is modeled as an m x n binary grid isInfected, where isInfected[i][j] == 0 represents uninfected cells, and isInfected[i][j] == 1 represents cells contaminated with the virus.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Array, Matrix, Simulation
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: isInfected = [[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:
On the second day, add 5 walls to quarantine the viral region on the right. The virus is fully contained.
Python Solution
class Solution:
def containVirus(self, isInfected: List[List[int]]) -> int:
def dfs(i, j):
vis[i][j] = True
areas[-1].append((i, j))
for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n:
if isInfected[x][y] == 1 and not vis[x][y]:
dfs(x, y)
elif isInfected[x][y] == 0:
c[-1] += 1
boundaries[-1].add((x, y))
m, n = len(isInfected), len(isInfected[0])
ans = 0
while 1:
vis = [[False] * n for _ in range(m)]
areas = []
c = []
boundaries = []
for i, row in enumerate(isInfected):
for j, v in enumerate(row):
if v == 1 and not vis[i][j]:
areas.append([])
boundaries.append(set())
c.append(0)
dfs(i, j)
if not areas:
break
idx = boundaries.index(max(boundaries, key=len))
ans += c[idx]
for k, area in enumerate(areas):
if k == idx:
for i, j in area:
isInfected[i][j] = -1
else:
for i, j in area:
for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and isInfected[x][y] == 0:
isInfected[x][y] = 1
return ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.