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Given the array favoriteCompanies where favoriteCompanies[i] is the list of favorites companies for the ith person (indexed from 0).

Return the indices of people whose list of favorite companies is not a subset of any other list of favorites companies. You must return the indices in increasing order.

Example 1:

Input: favoriteCompanies = [["leetcode","google","facebook"],["google","microsoft"],["google","facebook"],["google"],["amazon"]]
Output: [0,1,4]
Explanation:
Person with index=2 has favoriteCompanies[2]=["google","facebook"] which is a subset of favoriteCompanies[0]=["leetcode","google","facebook"] corresponding to the person with index 0.
Person with index=3 has favoriteCompanies[3]=["google"] which is a subset of favoriteCompanies[0]=["leetcode","google","facebook"] and favoriteCompanies[1]=["google","microsoft"].
Other lists of favorite companies are not a subset of another list, therefore, the answer is [0,1,4].


Example 2:

Input: favoriteCompanies = [["leetcode","google","facebook"],["leetcode","amazon"],["facebook","google"]]
Output: [0,1]
Explanation: In this case favoriteCompanies[2]=["facebook","google"] is a subset of favoriteCompanies[0]=["leetcode","google","facebook"], therefore, the answer is [0,1].


Example 3:

Input: favoriteCompanies = [["leetcode"],["google"],["facebook"],["amazon"]]
Output: [0,1,2,3]


Constraints:

• 1 <= favoriteCompanies.length <= 100
• 1 <= favoriteCompanies[i].length <= 500
• 1 <= favoriteCompanies[i][j].length <= 20
• All strings in favoriteCompanies[i] are distinct.
• All lists of favorite companies are distinct, that is, If we sort alphabetically each list then favoriteCompanies[i] != favoriteCompanies[j].
• All strings consist of lowercase English letters only.

## Solution: Hashtable

Time complexity: O(n*n*m)
Space complexity: O(n*m)
where n is the # of people, m is the # of companies

## C++

Use int as key to make it faster.

## C++

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