# 1125. Smallest Sufficient Team

In a project, you have a list of required skills `req_skills`, and a list of people. The `ith` person `people[i]` contains a list of skills that the person has.

Consider a sufficient team: a set of people such that for every required skill in `req_skills`, there is at least one person in the team who has that skill. We can represent these teams by the index of each person.

• For example, `team = [0, 1, 3]` represents the people with skills `people[0]``people[1]`, and `people[3]`.

Return  any sufficient team of the smallest possible size, represented by the index of each person. You may return the answer in any order.

It is guaranteed an answer exists.

Example 1:

``````Input: req_skills = ["java","nodejs","reactjs"], people = [["java"],["nodejs"],["nodejs","reactjs"]]
Output: [0,2]
``````

Example 2:

``````Input: req_skills = ["algorithms","math","java","reactjs","csharp","aws"], people = [["algorithms","math","java"],["algorithms","math","reactjs"],["java","csharp","aws"],["reactjs","csharp"],["csharp","math"],["aws","java"]]
Output: [1,2]
``````

Constraints:

• `1 <= req_skills.length <= 16`
• `1 <= req_skills[i].length <= 16`
• `req_skills[i]` consists of lowercase English letters.
• All the strings of `req_skills` are unique.
• `1 <= people.length <= 60`
• `0 <= people[i].length <= 16`
• `1 <= people[i][j].length <= 16`
• `people[i][j]` consists of lowercase English letters.
• All the strings of `people[i]` are unique.
• Every skill in `people[i]` is a skill in `req_skills`.
• It is guaranteed a sufficient team exists.

``````class Solution {
public:
vector<int> smallestSufficientTeam(vector<string>& req_skills, vector<vector<string>>& people) {
int n = req_skills.size();
unordered_map<int, vector<int>> dp(1 << n);
dp[0] = {};
unordered_map<string, int> skillMap;
for (int i = 0; i < n; ++i) {
skillMap[req_skills[i]] = i;
}
for (int i = 0; i < people.size(); ++i) {
int skill = 0;
for (string str : people[i]) {
skill |= 1 << skillMap[str];
}
for (auto a : dp) {
int cur = a.first | skill;
if (!dp.count(cur) || dp[cur].size() > 1 + dp[a.first].size()) {
dp[cur] = a.second;
dp[cur].push_back(i);
}
}
}
return dp[(1 << n) - 1];
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1125

https://leetcode.com/problems/smallest-sufficient-team/