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Problem solution · C++

Smallest Sufficient Team

Smallest Sufficient Team: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
41 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

Dynamic programming

For Smallest Sufficient Team, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 41 lines of C++ from the credited upstream file 1125.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, cached states.
  • 4 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 codeSmallest Sufficient Team · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  vector<int> smallestSufficientTeam(vector<string>& req_skills,                                     vector<vector<string>>& people) {    const int n = req_skills.size();    const int nSkills = 1 << n;    unordered_map<string, int> skillToId;    // dp[i] := the minimum people's indices to cover skillset of mask i    unordered_map<int, vector<int>> dp;    dp.reserve(nSkills);  // Avoid rehash.    dp[0] = {};     for (int i = 0; i < n; ++i)      skillToId[req_skills[i]] = i;     auto getSkill = [&](const vector<string>& person) {      int mask = 0;      for (const string& skill : person)        if (const auto it = skillToId.find(skill); it != skillToId.cend())          mask |= 1 << it->second;      return mask;    };     for (int i = 0; i < people.size(); ++i) {      const int currSkill = getSkill(people[i]);      for (const auto& [mask, indices] : dp) {        const int newSkillSet = mask | currSkill;        if (newSkillSet == mask)  // Adding people[i] doesn't increase skill set          continue;        if (!dp.contains(newSkillSet) ||            dp[newSkillSet].size() > indices.size() + 1) {          dp[newSkillSet] = indices;          dp[newSkillSet].push_back(i);        }      }    }     return dp[nSkills - 1];  }};

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