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

Description

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.

 

Solutions

Solution: Hash Table

• Time complexity: O(nlogn)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {string[][]} favoriteCompanies
 * @return {number[]}
 */
const peopleIndexes = function (favoriteCompanies) {
  const companiesSet = [];
  const sortFavorites = favoriteCompanies
    .map((favorites, index) => ({ favorites, index }))
    .sort((a, b) => b.favorites.length - a.favorites.length);

  return sortFavorites
    .reduce((result, { favorites, index }) => {
      const isSubset = companiesSet.some(companies => {
        return favorites.every(favorite => companies.has(favorite));
      });

      if (!isSubset) {
        companiesSet.push(new Set(favorites));
        result.push(index);
      }
      return result;
    }, [])
    .sort((a, b) => a - b);
};