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Problem solution · C++

Checking Existence of Edge Length Limited Paths II

Checking Existence of Edge Length Limited Paths II: 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
59 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 Checking Existence of Edge Length Limited Paths II, 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

  • 59 lines of C++ from the credited upstream file 1724.cpp.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 2 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 codeChecking Existence of Edge Length Limited Paths II · C++C++
Use this to learn the idea, then write your own version.
class UnionFind { public:  UnionFind() {}  UnionFind(int n) {    id.resize(n);     for (int i = 0; i < n; ++i)      id[i][0] = i;  }   void union_(int u, int v, int limit) {    const int i = find(u, limit);    const int j = find(v, limit);    if (i == j)      return;    id[i][limit] = j;  }   int find(int u, int limit) {    // the minimum iterator of id[u] > limit    const auto it = id[u].upper_bound(limit);    const int i = prev(it)->second;    if (i == u)      return u;    // Recursively find i's id.    const int j = find(i, limit);    id[u][limit] = j;    return j;  }  private:  // id[i]'s (key, value) := (limit, id of node i <= limit)  vector<map<int, int>> id;}; class DistanceLimitedPathsExist { public:  DistanceLimitedPathsExist(int n, vector<vector<int>>& edgeList) {    uf = UnionFind(n);     ranges::sort(edgeList, ranges::less{},                 [](const vector<int>& edge) { return edge[2]; });     for (const vector<int>& edge : edgeList) {      const int u = edge[0];      const int v = edge[1];      const int d = edge[2];      uf.union_(u, v, d);    }  }   bool query(int p, int q, int limit) {    return uf.find(p, limit - 1) == uf.find(q, limit - 1);  }  private:  UnionFind uf;};

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