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1697. Checking Existence of Edge Length Limited Paths 👍

  • Time: $O(n\log n + q\log q)$
  • Space: $(n + q)$
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class UnionFind {
 public:
  UnionFind(int n) : id(n) {
    iota(begin(id), end(id), 0);
  }

  void union_(int u, int v) {
    id[find(u)] = find(v);
  }

  int find(int u) {
    return id[u] == u ? u : id[u] = find(id[u]);
  }

 private:
  vector<int> id;
};

class Solution {
 public:
  vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList,
                                         vector<vector<int>>& queries) {
    vector<bool> ans(queries.size());
    UnionFind uf(n);

    for (int i = 0; i < queries.size(); ++i)
      queries[i].push_back(i);

    sort(begin(queries), end(queries),
         [](const auto& a, const auto& b) { return a[2] < b[2]; });
    sort(begin(edgeList), end(edgeList),
         [](const auto& a, const auto& b) { return a[2] < b[2]; });

    int i = 0;  // i := edgeList's index
    for (const vector<int>& q : queries) {
      // union edges whose distances < limit (q[2])
      while (i < edgeList.size() && edgeList[i][2] < q[2])
        uf.union_(edgeList[i][0], edgeList[i++][1]);
      if (uf.find(q[0]) == uf.find(q[1]))
        ans[q.back()] = true;
    }

    return ans;
  }
};
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class UnionFind {
  public UnionFind(int n) {
    id = new int[n];
    for (int i = 0; i < n; ++i)
      id[i] = i;
  }

  public void union(int u, int v) {
    id[find(u)] = find(v);
  }

  public int find(int u) {
    return id[u] == u ? u : (id[u] = find(id[u]));
  }

  private int[] id;
}

class Solution {
  public boolean[] distanceLimitedPathsExist(int n, int[][] edgeList, int[][] queries) {
    boolean[] ans = new boolean[queries.length];
    int[][] qs = new int[queries.length][4];
    UnionFind uf = new UnionFind(n);

    for (int i = 0; i < queries.length; ++i) {
      qs[i][0] = queries[i][0];
      qs[i][1] = queries[i][1];
      qs[i][2] = queries[i][2];
      qs[i][3] = i;
    }

    Arrays.sort(qs, (a, b) -> a[2] - b[2]);
    Arrays.sort(edgeList, (a, b) -> a[2] - b[2]);

    int i = 0; // i := edgeList's index
    for (int[] q : qs) {
      // union edges whose distances < limit (q[2])
      while (i < edgeList.length && edgeList[i][2] < q[2])
        uf.union(edgeList[i][0], edgeList[i++][1]);
      if (uf.find(q[0]) == uf.find(q[1]))
        ans[q[3]] = true;
    }

    return ans;
  }
}