1125. Smallest Sufficient Team

# Time:  O(m * 2^n), n is the number of skills
#                    m is the number of people
# Space: O(2^n)

class Solution(object):
    def smallestSufficientTeam(self, req_skills, people):
        """
        :type req_skills: List[str]
        :type people: List[List[str]]
        :rtype: List[int]
        """
        lookup = {v: i for i, v in enumerate(req_skills)}
        dp = {0: []}
        for i, p in enumerate(people):
            his_skill_set = 0
            for skill in p:
                if skill in lookup:
                    his_skill_set |= 1 << lookup[skill]
            for skill_set, people in dp.items():
                with_him = skill_set | his_skill_set
                if with_him == skill_set: continue
                if with_him not in dp or \
                   len(dp[with_him]) > len(people)+1:
                    dp[with_him] = people + [i]
        return dp[(1<<len(req_skills))-1]

Solution from kamyu104/LeetCode-Solutions · MIT