Smallest Sufficient Team

In a project, you have a list of required skills req_skills, and a list of people.  The i-th person people[i] contains a list of skills that 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 <= people.length <= 60
• 1 <= people[i].length, req_skills[i].length, people[i][j].length <= 16
• Elements of req_skills and people[i] are (respectively) distinct.
• req_skills[i][j], people[i][j][k] are lowercase English letters.
• Every skill in people[i] is a skill in req_skills.
• It is guaranteed a sufficient team exists.

思路：最近學到了這個強行剪枝的技巧，就是收集到的size > 目前的結果，直接return；這題的思路就是先把skill 和set of people建立好，然後去用skill set做backtracking收集，如果temp team的size大於結果，直接return，否則update結果，這裏有個小tricky的地方，就是如果people是新人，加入之後dfs，backtracking的時候，要判斷如果是新人，則remove，否則不remove；

class Solution {
    private int teamSize;
    private Set<Integer> resTeam;
    
    public int[] smallestSufficientTeam(String[] req_skills, List<List<String>> people) {
        int n = people.size();
        teamSize = n;
        resTeam = new HashSet<Integer>();
        HashMap<String, Set<Integer>> map = new HashMap<>();
        // map input to <Skill, Set<People>>;
        for(int i = 0; i < n; i++) {
            for(String skill: people.get(i)) {
                map.putIfAbsent(skill, new HashSet<Integer>());
                map.get(skill).add(i);
            }
        }
        // search res;
        HashSet<Integer> team = new HashSet<>();
        dfs(map, req_skills, team, 0);
        
        // collect res;
        int[] res = new int[resTeam.size()];
        int index = 0;
        for(int person: resTeam) {
            res[index++] = person;
        }
        return res;
    }
    
    private void dfs(HashMap<String, Set<Integer>> map, String[] req_skills, HashSet<Integer> team, int index) {
        if(team.size() > teamSize) {
            return;
        }
        if(index == req_skills.length) {
            teamSize = team.size();
            resTeam = new HashSet<Integer>(team);
            return;
        }
        for(int people: map.get(req_skills[index])) {
            boolean isNewPerson = !team.contains(people);
            team.add(people);
            dfs(map, req_skills, team, index + 1);
            if(isNewPerson) {
                team.remove(people);
            }
        }
    }
}