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Problem solution · Python

Minimize Maximum Component Cost

Minimize Maximum Component Cost: a Python solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Disjoint set union
Source
Kamyu LeetCode Solutions
Length
49 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 Minimize Maximum Component Cost, 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

  • 49 lines of Python from the credited upstream file minimize-maximum-component-cost.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 Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimize Maximum Component Cost · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n + eloge)# Space: O(n) # backward simulation, union find, sortclass UnionFind(object):  # Time: O(n * alpha(n)), Space: O(n)    def __init__(self, n):        self.set = range(n)        self.rank = [0]*n     def find_set(self, x):        stk = []        while self.set[x] != x:  # path compression            stk.append(x)            x = self.set[x]        while stk:            self.set[stk.pop()] = x        return x     def union_set(self, x, y):        x, y = self.find_set(x), self.find_set(y)        if x == y:            return False        if self.rank[x] > self.rank[y]:  # union by rank            x, y = y, x        self.set[x] = self.set[y]        if self.rank[x] == self.rank[y]:            self.rank[y] += 1        return True  class Solution(object):    def minCost(self, n, edges, k):        """        :type n: int        :type edges: List[List[int]]        :type k: int        :rtype: int        """        edges.sort(key=lambda x: x[2])        cnt = 0        uf = UnionFind(n)        for u, v, w in edges:            if not uf.union_set(u, v):                continue            cnt += 1            if cnt == n-k:                return w        return 0

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