Formatted question description: https://leetcode.ca/all/1452.html
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 <= 1001 <= favoriteCompanies[i].length <= 5001 <= 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 1.
// OJ: https://leetcode.com/problems/people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list/
// Time: O(NM + N^2) where N is the length of A and M is the maximum length of A[i]
// Space: O(NM)
class Solution {
public:
vector<int> peopleIndexes(vector<vector<string>>& A) {
int id = 0, N = A.size();
vector<bitset<50000>> v;
unordered_map<string, int> m;
for (auto &p : A) {
bitset<50000> bs;
for (auto &c : p) {
if (m.count(c) == 0) m[c] = id++;
bs.set(m[c]);
}
v.push_back(bs);
}
vector<int> ans;
for (int i = 0; i < N; ++i) {
int j = 0;
for (; j < N; ++j) {
if (i == j) continue;
if ((v[i] | v[j]) == v[j]) break;
}
if (j == N) ans.push_back(i);
}
return ans;
}
};
Java
class Solution {
public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
for (List<String> list : favoriteCompanies)
Collections.sort(list);
List<Integer> indices = new ArrayList<Integer>();
int peopleCount = favoriteCompanies.size();
for (int i = 0; i < peopleCount; i++) {
boolean flag = true;
List<String> list1 = favoriteCompanies.get(i);
for (int j = 0; j < peopleCount; j++) {
if (i == j)
continue;
List<String> list2 = favoriteCompanies.get(j);
if (isSubset(list1, list2)) {
flag = false;
break;
}
}
if (flag)
indices.add(i);
}
return indices;
}
public boolean isSubset(List<String> list1, List<String> list2) {
int size1 = list1.size(), size2 = list2.size();
if (size1 > size2)
return false;
int index1 = 0, index2 = 0;
while (index1 < size1 && index2 < size2) {
if (list1.get(index1).equals(list2.get(index2)))
index1++;
index2++;
}
return index1 == size1;
}
}