Skip to content

685. Redundant Connection II 👍

  • Time: $O(|V|)$
  • Space: $O(|V|)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
class UnionFind {
 public:
  UnionFind(int 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];
    }
    return true;
  }

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

class Solution {
 public:
  vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
    vector<int> ids(edges.size() + 1);
    int nodeWithTwoParents = 0;

    for (const vector<int>& edge : edges) {
      const int v = edge[1];
      if (++ids[v] == 2) {
        nodeWithTwoParents = v;
        break;
      }
    }

    // If there is no edge with two ids, don't skip any edge.
    if (nodeWithTwoParents == 0)
      return findRedundantDirectedConnection(edges, -1);

    for (int i = edges.size() - 1; i >= 0; --i)
      if (edges[i][1] == nodeWithTwoParents)
        // Try to delete the edges[i].
        if (findRedundantDirectedConnection(edges, i).empty())
          return edges[i];

    throw;
  }

  vector<int> findRedundantDirectedConnection(const vector<vector<int>>& edges,
                                              int skippedEdgeIndex) {
    UnionFind uf(edges.size() + 1);

    for (int i = 0; i < edges.size(); ++i) {
      if (i == skippedEdgeIndex)
        continue;
      if (!uf.unionByRank(edges[i][0], edges[i][1]))
        return edges[i];
    }

    return {};
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
class UnionFind {
  public UnionFind(int n) {
    id = new int[n];
    rank = new int[n];
    for (int i = 0; i < n; ++i)
      id[i] = i;
  }

  public boolean unionByRank(int u, int v) {
    final int i = find(u);
    final 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];
    }
    return true;
  }

  private int[] id;
  private int[] rank;

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

class Solution {
  public int[] findRedundantDirectedConnection(int[][] edges) {
    int[] ids = new int[edges.length + 1];
    int nodeWithTwoParents = 0;

    for (int[] edge : edges) {
      final int v = edge[1];
      if (++ids[v] == 2) {
        nodeWithTwoParents = v;
        break;
      }
    }

    // If there is no edge with two ids, don't skip any edge.
    if (nodeWithTwoParents == 0)
      return findRedundantDirectedConnection(edges, -1);

    for (int i = edges.length - 1; i >= 0; --i)
      if (edges[i][1] == nodeWithTwoParents)
        // Try to delete the edges[i].
        if (findRedundantDirectedConnection(edges, i).length == 0)
          return edges[i];

    throw new IllegalArgumentException();
  }

  private int[] findRedundantDirectedConnection(int[][] edges, int skippedEdgeIndex) {
    UnionFind uf = new UnionFind(edges.length + 1);

    for (int i = 0; i < edges.length; ++i) {
      if (i == skippedEdgeIndex)
        continue;
      if (!uf.unionByRank(edges[i][0], edges[i][1]))
        return edges[i];
    }

    return new int[] {};
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
class UnionFind:
  def __init__(self, n: int):
    self.id = list(range(n))
    self.rank = [0] * n

  def unionByRank(self, u: int, v: int) -> bool:
    i = self._find(u)
    j = self._find(v)
    if i == j:
      return False
    if self.rank[i] < self.rank[j]:
      self.id[i] = j
    elif self.rank[i] > self.rank[j]:
      self.id[j] = i
    else:
      self.id[i] = j
      self.rank[j] += 1
    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 findRedundantDirectedConnection(
      self, edges: list[list[int]],
  ) -> list[int]:
    ids = [0] * (len(edges) + 1)
    nodeWithTwoParents = 0

    for _, v in edges:
      ids[v] += 1
      if ids[v] == 2:
        nodeWithTwoParents = v

    def findRedundantDirectedConnection(skippedEdgeIndex: int) -> list[int]:
      uf = UnionFind(len(edges) + 1)

      for i, edge in enumerate(edges):
        if i == skippedEdgeIndex:
          continue
        if not uf.unionByRank(edge[0], edge[1]):
          return edge

      return []

    # If there is no edge with two ids, don't skip any edge.
    if nodeWithTwoParents == 0:
      return findRedundantDirectedConnection(-1)

    for i in reversed(range(len(edges))):
      _, v = edges[i]
      if v == nodeWithTwoParents:
        # Try to delete the edges[i].
        if not findRedundantDirectedConnection(i):
          return edges[i]