human
What is a the Python code you requested given the following:
Solve in Python:
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.
assistant
#Here is the Python code you requested.

from functools import lru_cache

class Solution:
    def smallestSufficientTeam(self, req_skills: List[str], people: List[List[str]]) -> List[int]:
        ids = {}
        sid = 0
        skills = 0
        for skill in req_skills:
            ids[skill] = sid
            skills = skills | (1 << sid)
            # print(sid, 1 << sid)
            sid += 1

        # print(ids)
        # print(skills)

        n = len(people)

        @lru_cache(None)
        def dp(i, left):
            if left == 0:
                return []
            if i == n:
                return None

            a = dp(i + 1, left)

            for s in people[i]:
                if s in ids:
                    j = ids[s]
                    left = (left & ~(1 << j))

            b = dp(i + 1, left)

            if a is not None and b is not None:
                if len(a) < len(b) + 1:
                    return a
                else:
                    return [i] + b

            if a is None and b is None:
                return None

            return ([i] + b) if a is None else a

        return dp(0, skills)