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LeetCode Solutions

684. Redundant Connection

Time: $O(n\log n)$

Space: $O(n)$

			


class UnionFind {
 public:
  UnionFind(int n) : id(n) {
    iota(begin(id), end(id), 0);
  }

  bool union_(int u, int v) {
    const int i = find(u);
    const int j = find(v);
    if (i == j)
      return false;
    id[i] = j;
    return true;
  }

 private:
  vector<int> id;

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

class Solution {
 public:
  vector<int> findRedundantConnection(vector<vector<int>>& edges) {
    UnionFind uf(edges.size() + 1);

    for (const vector<int>& e : edges)
      if (!uf.union_(e[0], e[1]))
        return e;

    throw;
  }
};




			


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

  public boolean union(int u, int v) {
    final int i = find(u);
    final int j = find(v);
    if (i == j)
      return false;
    id[i] = j;
    return true;
  }

  private int[] id;

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

class Solution {
  public int[] findRedundantConnection(int[][] edges) {
    UnionFind uf = new UnionFind(edges.length + 1);

    for (int[] e : edges)
      if (!uf.union(e[0], e[1]))
        return edge;

    throw new IllegalArgumentException();
  }
}




			


class UnionFind:
  def __init__(self, n: int):
    self.id = [i for i in range(n + 1)]

  def union(self, u: int, v: int) -> bool:
    i = self.find(u)
    j = self.find(v)
    if i == j:
      return False
    self.id[i] = j
    return True

  def find(self, u: int) -> int:
    if self.id[u] != u:
      self.id[u] = self.find(self.id[u])
    return self.id[u]


class Solution:
  def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
    uf = UnionFind(len(edges))

    for edge in edges:
      if not uf.union(edge[0], edge[1]):
        return edge