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Problem solution · Python

Redundant Connection II

Redundant Connection II: a Python 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 Redundant Connection 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 Python from the credited upstream file 685.py.
  • The implementation visibly relies on sequence storage.
  • No explicit 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 codeRedundant Connection II · PythonPython
Use this to learn the idea, then write your own version.
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]

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