leetcode - 1125. Smallest Sufficient Team

Description

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.

Solution

Solved after reading others’ solutions

Use bits to denote skills:
1<<0 to denote skills[0]
1<<1 to denote skills[1]
…

We use dp[skills] to denote the minimum number of people that have these skills, then iterate people, for every new people, use map to get their skills, then iterate dp, new skills would be skills | new_skills_from_people, if new skills is not in dp, append directly. Otherwise compare the length and update using the shorter one.

Time complexity: $o(people\ast 2^{skills})$
Space complexity: $o(2^{skills})$

Code

class Solution:
    def smallestSufficientTeam(self, req_skills: List[str], people: List[List[str]]) -> List[int]:
        dp = {0: []}
        skill_map = {each_skill: i for i, each_skill in enumerate(req_skills)}
        all_people = list(range(len(people)))
        for people_index, people_skills in enumerate(people):
            skill_bit = 0
            for item in people_skills:
                skill_bit |= (1 << skill_map[item])
            old_dp = dp.copy()
            for each_skill_bit, people_in_dp in old_dp.items():
                new_skill_set = each_skill_bit | skill_bit
                if len(dp.get(new_skill_set, all_people)) > len(people_in_dp) + 1:
                    dp[new_skill_set] = people_in_dp + [people_index]
        all_skill = 0
        for skill in skill_map.values():
            all_skill |= (1 << skill)
        return dp[all_skill]