Solution Explanation

This problem asks to find the indices of people whose favorite company lists are not subsets of any other person's list. The most efficient approach uses a hash table (or map in other languages) to manage company names and sets to represent each person's favorite companies.

Algorithm:

1. Create a mapping: We first create a mapping (dictionary in Python, HashMap in Java, etc.) to assign a unique integer ID to each distinct company name. This allows us to represent the sets of favorite companies efficiently using integers.

2. Convert to sets: We convert each person's list of favorite companies into a set of integer IDs (using the mapping created in step 1). Sets are used because they provide efficient membership testing, which is crucial for the subset check.

3. Subset check: For each person, we iterate through all other people. If a person's favorite company set is a subset of another person's set, then that person's index is not added to the result.

4. Collect and return: Finally, we collect the indices of people whose sets were not subsets of any others and return them in increasing order.

Time Complexity Analysis:

• Step 1: Creating the company-to-ID mapping takes O(C) time, where C is the total number of unique company names across all lists.

• Step 2: Converting each person's list to a set takes O(P * M) time in the worst case, where P is the number of people and M is the maximum number of companies in a single person's list.

• Step 3: The subset check involves nested loops, iterating through all pairs of people. For each pair, checking if one set is a subset of another takes O(M) time in the worst case. Thus, this step takes O(P² * M) time.

• Step 4: Collecting and returning the result takes O(P) time.

Therefore, the overall time complexity is dominated by the subset check, resulting in O(P² * M), where P is the number of people and M is the maximum number of companies a person favors.

Space Complexity Analysis:

• Company mapping: The space required for the company-to-ID mapping is O(C), where C is the total number of unique companies.

• Sets: The space for storing the sets of favorite companies is O(P * M), where P is the number of people and M is the maximum number of companies in a person's list.

Therefore, the overall space complexity is O(P * M + C). Since C is bounded by P * M, we can simplify this to O(P * M).

Code Examples (with minor improvements for clarity):

The provided code examples are already fairly efficient, but here's a slightly modified Python version with more concise subset checking:

from typing import List
 
class Solution:
    def peopleIndexes(self, favoriteCompanies: List[List[str]]) -> List[int]:
        n = len(favoriteCompanies)
        company_map = {}
        next_id = 0
        
        # Build company map
        for companies in favoriteCompanies:
            for company in companies:
                if company not in company_map:
                    company_map[company] = next_id
                    next_id += 1
        
        # Convert to sets of IDs
        company_sets = []
        for companies in favoriteCompanies:
            company_sets.append(set(company_map[c] for c in companies))
 
        result = []
        for i in range(n):
            is_subset = False
            for j in range(n):
                if i != j and company_sets[i].issubset(company_sets[j]):
                    is_subset = True
                    break
            if not is_subset:
                result.append(i)
        return result
 

The other language examples (Java, C++, Go, TypeScript) follow similar logic with appropriate data structures for each language. They all achieve the same time and space complexity as described above.