Problem solution · C++

Minimize Maximum Component Cost

Minimize Maximum Component Cost: a C++ 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
60 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

  • 60 lines of C++ from the credited upstream file minimize-maximum-component-cost.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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.

Full codeMinimize Maximum Component Cost · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n + eloge)// Space: O(n) // backward simulation, union find, sortclass Solution {public:    int minCost(int n, vector<vector<int>>& edges, int k) {        sort(begin(edges), end(edges), [](const auto& a, const auto& b) {            return a[2] < b[2];        });        int cnt = 0;        UnionFind uf(n);        for (const auto& e : edges) {            if (!uf.union_set(e[0], e[1])) {                continue;            }            if (++cnt == n - k) {                return e[2];            }        }        return 0;    } private:    class UnionFind {    public:        UnionFind(int n)         : set_(n)         , rank_(n) {            iota(set_.begin(), set_.end(), 0);        }         int find_set(int x) {           if (set_[x] != x) {               set_[x] = find_set(set_[x]);  // Path compression.           }           return set_[x];        }         bool union_set(int x, int y) {            x = find_set(x), y = find_set(y);            if (x == y) {                return false;            }            if (rank_[x] > rank_[y]) {                swap(x, y);            }            set_[x] = y;  // Union by rank.            if (rank_[x] == rank_[y]) {                ++rank_[y];            }            return true;        }     private:        vector<int> set_;        vector<int> rank_;    };}; 

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