1452. People Whose List of Favorite Companies Is Not a Subset of Another List

Explanation

To solve this problem, we can iterate over each person's list of favorite companies and check if it is a subset of any other person's list. We can do this by comparing each person's list with all other lists. If a person's list is not a subset of any other list, we add that person's index to the result.

Create a set for each person's list of favorite companies.
Iterate over each person's list and compare it with all other lists.
If a person's list is a subset of any other list, mark that person's index.
Return the indices of people whose list of favorite companies is not a subset of any other list.

Time Complexity: O(n^2 * m), where n is the number of people and m is the average length of a list of favorite companies. Space Complexity: O(n * m), where n is the number of people and m is the average length of a list of favorite companies.

View Leetcode Problem

import java.util.*;

class Solution {
    public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
        List<Integer> result = new ArrayList<>();
        Set<String>[] sets = new HashSet[favoriteCompanies.size()];

        for (int i = 0; i < favoriteCompanies.size(); i++) {
            sets[i] = new HashSet<>(favoriteCompanies.get(i));
        }

        for (int i = 0; i < favoriteCompanies.size(); i++) {
            boolean isSubset = false;
            for (int j = 0; j < favoriteCompanies.size(); j++) {
                if (i != j && sets[j].containsAll(sets[i])) {
                    isSubset = true;
                    break;
                }
            }
            if (!isSubset) {
                result.add(i);
            }
        }

        return result;
    }
}

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Explanation

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