CPMagical CP Library

Problem solution · Python

Minimum Weighted Subgraph With the Required Paths II

Minimum Weighted Subgraph With the Required Paths II: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
82 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

Direct simulation

For Minimum Weighted Subgraph With the Required Paths II, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 82 lines of Python from the credited upstream file 3553.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 Weighted Subgraph With the Required Paths II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  # Similar to 2846. Minimum Edge Weight Equilibrium Queries in a Tree  def minimumWeight(      self,      edges: list[list[int]],      queries: list[list[int]]  ) -> list[int]:    n = len(edges) + 1    m = math.ceil(math.log2(n))    graph = [[] for _ in range(n)]    jump = [[0] * m for _ in range(n)]  # jump[i][j] := the 2^j-th ancestor of i    depth = [0] * n  # depth[i] := the depth of i    dist = [0] * n  # dist[i] := the distance from root to i     for u, v, w in edges:      graph[u].append((v, w))      graph[v].append((u, w))     self._dfs(graph, 0, -1, jump, depth, dist)     for j in range(1, m):      for i in range(n):        jump[i][j] = jump[jump[i][j - 1]][j - 1]     return [(self._distance(src1, src2, jump, depth, dist) +             self._distance(src1, dest, jump, depth, dist) +             self._distance(src2, dest, jump, depth, dist)) // 2            for src1, src2, dest in queries]   def _dfs(      self,      graph: list[list[tuple[int, int]]],      u: int,      prev: int,      jump: list[list[int]],      depth: list[int],      dist: list[int]  ) -> None:    for v, w in graph[u]:      if v == prev:        continue      jump[v][0] = u      depth[v] = depth[u] + 1      dist[v] = dist[u] + w      self._dfs(graph, v, u, jump, depth, dist)   def _getLCA(      self,      u: int,      v: int,      jump: list[list[int]],      depth: list[int]  ) -> int:    """Returns the lca(u, v) by binary jump."""    # v is always deeper than u.    if depth[u] > depth[v]:      return self._getLCA(v, u, jump, depth)    # Jump v to the same height of u.    for j in range(len(jump[0])):      if depth[v] - depth[u] >> j & 1:        v = jump[v][j]    if u == v:      return u    # Jump u and v to the node right below the lca.    for j in range(len(jump[0]) - 1, -1, -1):      if jump[u][j] != jump[v][j]:        u = jump[u][j]        v = jump[v][j]    return jump[u][0]   def _distance(      self,      u: int,      v: int,      jump: list[list[int]],      depth: list[int],      dist: list[int]  ) -> int:    """Returns the distance between u and v."""    lca = self._getLCA(u, v, jump, depth)    return dist[u] + dist[v] - 2 * dist[lca]

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 ↗