CPMagical CP Library

Problem solution · Python

Minimum Cost Walk in Weighted Graph

Minimum Cost Walk in Weighted Graph: 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
50 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 Minimum Cost Walk in Weighted Graph, 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

  • 50 lines of Python from the credited upstream file 3108.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 codeMinimum Cost Walk in Weighted Graph · 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    # 2^17 - 1 is the minimum number in the form 2^x - 1 > 10^5.    self.weight = [(1 << 17) - 1] * n   def unionByRank(self, u: int, v: int, w: int) -> None:    i = self._find(u)    j = self._find(v)    newWeight = self.weight[i] & self.weight[j] & w    self.weight[i] = newWeight    self.weight[j] = newWeight    if i == j:      return    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   def getMinCost(self, u: int, v: int) -> int:    if u == v:      return 0    i = self._find(u)    j = self._find(v)    return self.weight[i] if i == j else -1   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 minimumCost(      self,      n: int,      edges: list[list[int]],      query: list[list[int]],  ) -> list[int]:    uf = UnionFind(n)     for u, v, w in edges:      uf.unionByRank(u, v, w)     return [uf.getMinCost(u, v) for u, v in query]

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗