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

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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:

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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:

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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:

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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.

Hints/Notes

  • 2025/03/14 Q1
  • bitset
  • No solution from 0x3F or Leetcode

Solution

Language: C++

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class Solution {
public:
    vector<int> peopleIndexes(vector<vector<string>>& favoriteCompanies) {
        unordered_set<string> s;
        for (auto& companies : favoriteCompanies) {
            for (auto& c : companies) {
                s.insert(c);
            }
        }
        int idx = 0;
        unordered_map<string, int> companyToIndex;
        for (string c : s) {
            companyToIndex[c] = idx++;
        }
        int numPeople = favoriteCompanies.size();
        vector<bitset<50000>> people(numPeople);
        for (int i = 0; i < numPeople; i++) {
            for (auto& c : favoriteCompanies[i]) {
                people[i].set(companyToIndex[c]);
            }
        }
        vector<int> res;
        for (int i = 0; i < numPeople; i++) {
            bool valid = true;
            for (int j = 0; j < numPeople; j++) {
                if (i == j) continue;
                bitset<50000> intersection = people[i] & people[j];
                if (intersection.count() == people[i].count()) {
                    valid = false;
                    break;
                }
            }
            if (valid) res.push_back(i);
        }
        return res;
    }
};