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Problem solution · C++

Maximum Points Activated with One Addition

Maximum Points Activated with One Addition: 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
84 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 Maximum Points Activated with One Addition, 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

  • 84 lines of C++ from the credited upstream file maximum-points-activated-with-one-addition.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 6 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 codeMaximum Points Activated with One Addition · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n)// Space: O(n) // union findclass Solution {public:    int maxActivated(vector<vector<int>>& points) {        static const int N_DIM = 2;        static const int N_ADD = 1;         UnionFind uf(size(points));        vector<unordered_map<int, int>> lookup(N_DIM);        for (int i = 0; i < size(points); ++i) {            for (int j = 0; j < size(lookup); ++j) {                if (lookup[j].count(points[i][j])) {                    uf.union_set(i, lookup[j][points[i][j]]);                } else {                    lookup[j][points[i][j]] = i;                }            }        }        vector<int> top(min(N_ADD + 1, static_cast<int>(size(points))));        for (int i = 0; i < size(points); ++i) {            if (uf.find_set(i) != i) {                continue;            }            for (int j = 0, s = uf.total(i); j < size(top); ++j) {                if (s > top[j]) {                    swap(top[j], s);                }            }        }        return accumulate(cbegin(top), cend(top), 0) + N_ADD;     } private:    class UnionFind {    public:        UnionFind(int n)         : set_(n)         , rank_(n)         , size_(n, 1)  {            iota(set_.begin(), set_.end(), 0);        }         int find_set(int x) {           vector<int> stk;            while (set_[x] != x) {  // path compression                stk.emplace_back(x);                x = set_[x];            }            while (!empty(stk)) {                const int y = stk.back(); stk.pop_back();                set_[y] = x;            }            return 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);            } else if (rank_[x] == rank_[y]) {                ++rank_[y];            }            set_[x] = y;  // Union by rank.            size_[y] += size_[x];            return true;        }         int64_t total(int x) {            return size_[find_set(x)];        }     private:        vector<int> set_;        vector<int> rank_;        vector<int64_t> size_;    };};

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