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Problem solution · C++

Remove Max Number of Edges to Keep Graph Fully Traversable

Remove Max Number of Edges to Keep Graph Fully Traversable: a C++ solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Disjoint set union
Source
walkccc LeetCode Solutions
Length
77 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

Disjoint set union

For Remove Max Number of Edges to Keep Graph Fully Traversable, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 77 lines of C++ from the credited upstream file 1579.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

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 codeRemove Max Number of Edges to Keep Graph Fully Traversable · C++C++
Use this to learn the idea, then write your own version.
class UnionFind { public:  UnionFind(int n) : count(n), id(n), rank(n) {    iota(id.begin(), id.end(), 0);  }   bool unionByRank(int u, int v) {    const int i = find(u);    const int j = find(v);    if (i == j)      return false;    if (rank[i] < rank[j]) {      id[i] = j;    } else if (rank[i] > rank[j]) {      id[j] = i;    } else {      id[i] = j;      ++rank[j];    }    --count;    return true;  }   int getCount() const {    return count;  }  private:  int count;  vector<int> id;  vector<int> rank;   int find(int u) {    return id[u] == u ? u : id[u] = find(id[u]);  }}; class Solution { public:  int maxNumEdgesToRemove(int n, vector<vector<int>>& edges) {    UnionFind alice(n);    UnionFind bob(n);    int requiredEdges = 0;     // Greedily put type 3 edges in the front.    ranges::sort(edges, ranges::greater{},                 [](const vector<int>& edge) { return edge[0]; });     for (const vector<int>& edge : edges) {      const int type = edge[0];      const int u = edge[1] - 1;      const int v = edge[2] - 1;      switch (type) {        case 3:  // Can be traversed by Alice and Bob.          // Note that we should use | instead of || because if the first          // expression is true, short-circuiting will skip the second          // expression.          if (alice.unionByRank(u, v) | bob.unionByRank(u, v))            ++requiredEdges;          break;        case 2:  // Can be traversed by Bob.          if (bob.unionByRank(u, v))            ++requiredEdges;          break;        case 1:  // Can be traversed by Alice.          if (alice.unionByRank(u, v))            ++requiredEdges;          break;      }    }     return alice.getCount() == 1 && bob.getCount() == 1               ? edges.size() - requiredEdges               : -1;  }};

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