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Problem solution · C++

Contain Virus

Contain Virus: a C++ solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
86 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Contain Virus, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 86 lines of C++ from the credited upstream file 749.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 6 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeContain Virus · C++C++
Use this to learn the idea, then write your own version.
struct Region {  // Given m = the number of rows and n = the number of columns, (x, y) will be  // hashed as x * n + y.  unordered_set<int> infected;  unordered_set<int> noninfected;  int wallsRequired = 0;}; class Solution { public:  int containVirus(vector<vector<int>>& isInfected) {    const int m = isInfected.size();    const int n = isInfected[0].size();    int ans = 0;     while (true) {      vector<Region> regions;      vector<vector<bool>> seen(m, vector<bool>(n));       for (int i = 0; i < m; ++i)        for (int j = 0; j < n; ++j)          if (isInfected[i][j] == 1 && !seen[i][j]) {            Region region;            // Use DFS to find all the regions (1s).            dfs(isInfected, i, j, region, seen);            if (!region.noninfected.empty())              regions.push_back(region);          }       if (regions.empty())        break;  // No region causes further infection.       // Regions that infect the most neighbors will be sorted to the back of      // the array.      ranges::sort(regions, ranges::less{}, [](const Region& region) {        return region.noninfected.size();      });       // Build walls around the region that infects the most neighbors.      Region mostInfectedRegion = regions.back();      regions.pop_back();      ans += mostInfectedRegion.wallsRequired;       for (const int neighbor : mostInfectedRegion.infected) {        const int i = neighbor / n;        const int j = neighbor % n;        // The isInfected is now contained and won't be infected anymore.        isInfected[i][j] = 2;      }       // For remaining regions, infect their neighbors.      for (const Region& region : regions)        for (const int neighbor : region.noninfected) {          const int i = neighbor / n;          const int j = neighbor % n;          isInfected[i][j] = 1;        }    }     return ans;  }  private:  void dfs(const vector<vector<int>>& isInfected, int i, int j, Region& region,           vector<vector<bool>>& seen) {    if (i < 0 || i == isInfected.size() || j < 0 || j == isInfected[0].size())      return;    if (seen[i][j] || isInfected[i][j] == 2)      return;    if (isInfected[i][j] == 0) {      region.noninfected.insert(i * isInfected[0].size() + j);      ++region.wallsRequired;      return;    }     // isInfected[i][j] == 1    seen[i][j] = true;    region.infected.insert(i * isInfected[0].size() + j);     dfs(isInfected, i + 1, j, region, seen);    dfs(isInfected, i - 1, j, region, seen);    dfs(isInfected, i, j + 1, region, seen);    dfs(isInfected, i, j - 1, region, seen);  }};

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