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Problem solution · C++

Find Critical and Pseudo-Critical Edges in Minimum Spanning Tree

Find Critical and Pseudo-Critical Edges in Minimum Spanning Tree: 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
94 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 Find Critical and Pseudo-Critical Edges in Minimum Spanning Tree, 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

  • 94 lines of C++ from the credited upstream file 1489.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks 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 codeFind Critical and Pseudo-Critical Edges in Minimum Spanning Tree · C++C++
Use this to learn the idea, then write your own version.
class UnionFind { public:  UnionFind(int n) : id(n), rank(n) {    iota(id.begin(), id.end(), 0);  }   void unionByRank(int u, int v) {    const int i = find(u);    const int j = find(v);    if (i == j)      return;    if (rank[i] < rank[j]) {      id[i] = j;    } else if (rank[i] > rank[j]) {      id[j] = i;    } else {      id[i] = j;      ++rank[j];    }  }   int find(int u) {    return id[u] == u ? u : id[u] = find(id[u]);  }  private:  vector<int> id;  vector<int> rank;}; class Solution { public:  vector<vector<int>> findCriticalAndPseudoCriticalEdges(      int n, vector<vector<int>>& edges) {    vector<int> criticalEdges;    vector<int> pseudoCriticalEdges;     // Record the index information, so edges[i] := (u, v, weight, index).    for (int i = 0; i < edges.size(); ++i)      edges[i].push_back(i);     // Sort by the weight.    ranges::sort(edges, ranges::less{},                 [](const vector<int>& edge) { return edge[2]; });     const int mstWeight = getMSTWeight(n, edges, {}, -1);     for (const vector<int>& edge : edges) {      const int index = edge[3];      // Deleting the `edge` increases the MST's weight or makes the MST      // invalid.      if (getMSTWeight(n, edges, {}, index) > mstWeight)        criticalEdges.push_back(index);      // If an edge can be in any MST, we can always add `edge` to the edge set.      else if (getMSTWeight(n, edges, edge, -1) == mstWeight)        pseudoCriticalEdges.push_back(index);    }     return {criticalEdges, pseudoCriticalEdges};  }  private:  int getMSTWeight(int n, const vector<vector<int>>& edges,                   const vector<int>& firstEdge, int deletedEdgeIndex) {    int mstWeight = 0;    UnionFind uf(n);     if (!firstEdge.empty()) {      uf.unionByRank(firstEdge[0], firstEdge[1]);      mstWeight += firstEdge[2];    }     for (const vector<int>& edge : edges) {      const int u = edge[0];      const int v = edge[1];      const int weight = edge[2];      const int index = edge[3];      if (index == deletedEdgeIndex)        continue;      if (uf.find(u) == uf.find(v))        continue;      uf.unionByRank(u, v);      mstWeight += weight;    }     const int root = uf.find(0);    for (int i = 0; i < n; ++i)      if (uf.find(i) != root)        return INT_MAX;     return mstWeight;  }};

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