Class Solution


  • public class Solution
    extends Object
    1452 - People Whose List of Favorite Companies Is Not a Subset of Another List.

    Medium

    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.
    • Constructor Detail

      • Solution

        public Solution()