Naive Approach (Brute Force)

The most straightforward approach is to generate all possible subsets of people and check if each subset forms a sufficient team. Then, find the smallest such team.

1. Generate all subsets: Create all possible combinations of people indices.
2. Check sufficiency: For each subset, check if it covers all required skills.
3. Find the minimum: Keep track of the smallest sufficient team found so far.

Code (Python)

from itertools import combinations

def smallestSufficientTeam_brute_force(req_skills, people):
    skill_map = {skill: i for i, skill in enumerate(req_skills)}
    n_skills = len(req_skills)
    
    people_masks = []
    for person_skills in people:
        mask = 0
        for skill in person_skills:
            mask |= (1 << skill_map[skill])
        people_masks.append(mask)
    
    for team_size in range(1, len(people) + 1):
        for team in combinations(range(len(people)), team_size):
            covered_mask = 0
            for person_index in team:
                covered_mask |= people_masks[person_index]
            if covered_mask == (1 << n_skills) - 1:
                return list(team)
    return []

Time Complexity

• Generating all subsets takes O(2^P) time, where P is the number of people.
• Checking each subset's sufficiency takes O(N), where N is the number of required skills. In the worst case, it can go through all people.
• Overall: O(2^P * N), which is exponential.

Space Complexity

• O(P) to store the subset.
• O(N) for skill map.
• O(P) to store the people masks.
• Overall: O(P + N)

Optimal Approach (Dynamic Programming with Bit Masking)

We can significantly improve the efficiency by using dynamic programming with bit masking. The idea is to represent the covered skills as a bitmask. Each bit in the mask corresponds to a skill. If the i-th bit is set, it means the i-th skill is covered.

1. Skill Mapping: Create a mapping from skills to bit positions.
2. People Masks: Create a bitmask for each person representing the skills they possess.
3. DP State: dp[mask] stores the smallest team required to cover the skills represented by the bitmask mask.
4. DP Transition: Iterate through each person. For each person, update the dp array. If adding the person to the current team results in a smaller team size for a given covered skill set, update the dp array.

Code (Python)

def smallestSufficientTeam(req_skills, people):
    n = len(req_skills)
    skill_id = {s: i for i, s in enumerate(req_skills)}

    dp = {0: []}
    for i, p in enumerate(people):
        skills_mask = 0
        for skill in p:
            skills_mask |= 1 << skill_id[skill]

        for mask, team in list(dp.items()):
            new_mask = mask | skills_mask
            if new_mask not in dp or len(team) + 1 < len(dp[new_mask]):
                dp[new_mask] = team + [i]

    return dp[(1 << n) - 1]

Time Complexity

• O(P * 2^N), where P is the number of people and N is the number of required skills. We iterate through each person and each possible mask.

Space Complexity

• O(2^N) to store the dp array, where N is the number of required skills.

Edge Cases

• Empty req_skills: If req_skills is empty, any empty team will be a sufficient team, so return an empty list.
• Empty people: If people is empty and req_skills is not empty, then there is no sufficient team which violates the constraints. The problem states it is guaranteed a sufficient team exists, so this edge case does not need to be handled.
• Duplicate skills within a person's skills: The problem statement specifies that all skills of a person are unique, therefore no edge case to handle.