CPMagical CP Library

Problem solution · C++

Frequencies of Shortest Supersequences

Frequencies of Shortest Supersequences: a C++ solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
101 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

Hash-based lookup

For Frequencies of Shortest Supersequences, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 101 lines of C++ from the credited upstream file 3435.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 12 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 codeFrequencies of Shortest Supersequences · C++C++
Use this to learn the idea, then write your own version.
enum class State { kInit, kVisiting, kVisited }; class Solution { public:  vector<vector<int>> supersequences(vector<string>& words) {    vector<vector<int>> ans;    const vector<pair<int, int>> edges = getEdges(words);    const vector<int> nodes = getNodes(edges);    const vector<int> letterToIndex = getLetterToIndex(nodes);    vector<vector<int>> graph(nodes.size());     for (const auto& [u, v] : edges)      graph[letterToIndex[u]].push_back(letterToIndex[v]);     for (const auto& doubledSubset : getMinimumSubsets(graph)) {      vector<int> freq(26);      for (const int letter : nodes)        freq[letter] = 1;      for (const int index : doubledSubset)        freq[nodes[index]] = 2;      ans.push_back(freq);    }     return ans;  }  private:  // Returns a list of the minimum subsets of nodes that do not create a cycle  // when skipped.  vector<vector<int>> getMinimumSubsets(const vector<vector<int>>& graph) {    const int n = graph.size();    vector<vector<int>> res;    for (int subsetSize = 0; subsetSize <= n; ++subsetSize) {      vector<bool> combination(n);      fill(combination.end() - subsetSize, combination.end(), true);      do {        vector<int> doubledSubset;        for (int i = 0; i < n; ++i)          if (combination[i])            doubledSubset.push_back(i);        if (!hasCycleSkipping(graph,                              {doubledSubset.begin(), doubledSubset.end()}))          res.push_back(doubledSubset);      } while (next_permutation(combination.begin(), combination.end()));      if (!res.empty())        return res;    }    return res;  }   // Returns true if there is a cycle in the `graph` when skipping any edges  // whose both endpoints are in `doubledSubset`.  bool hasCycleSkipping(const vector<vector<int>>& graph,                        const unordered_set<int>& doubledSubset) {    vector<State> states(graph.size());    for (int i = 0; i < graph.size(); ++i)      if (hasCycle(graph, i, states, doubledSubset))        return true;    return false;  }   bool hasCycle(const vector<vector<int>>& graph, int u, vector<State>& states,                const unordered_set<int>& doubledSubset) {    if (states[u] == State::kVisiting)      return true;    if (states[u] == State::kVisited)      return false;    states[u] = State::kVisiting;    if (!doubledSubset.contains(u))      for (const int v : graph[u])        if (!doubledSubset.contains(v) &&            hasCycle(graph, v, states, doubledSubset))          return true;    states[u] = State::kVisited;    return false;  }   vector<pair<int, int>> getEdges(const vector<string>& words) {    vector<pair<int, int>> edges;    for (const string& word : words)      edges.push_back({word[0] - 'a', word[1] - 'a'});    return edges;  }   vector<int> getNodes(const vector<pair<int, int>>& edges) {    set<int> nodes;    for (const auto& [u, v] : edges) {      nodes.insert(u);      nodes.insert(v);    }    return {nodes.begin(), nodes.end()};  }   vector<int> getLetterToIndex(const vector<int>& nodes) {    vector<int> letterToIndex(26);    for (int i = 0; i < nodes.size(); ++i)      letterToIndex[nodes[i]] = i;    return letterToIndex;  }};

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 ↗