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

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.

收藏清单。

给你一个数组 favoriteCompanies ，其中 favoriteCompanies[i] 是第 i 名用户收藏的公司清单（下标从 0 开始）。

请找出不是其他任何人收藏的公司清单的子集的收藏清单，并返回该清单下标。下标需要按升序排列。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list
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第二次做周赛的题，我目前只能想到一个思路，代码可以通过，如果有更好的思路我之后再补充。首先我写了一个helper函数帮助判断某个list是不是另一个list的subset，有了这个辅助函数之后，我再遍历input list，然后再每两个list之间比较，是否较短的那个list是较长的那个list的subset，如果是就把较短的list的index加入hashset，说明这个较短的list是别人的subset。再次遍历input的index，如果index不在hashset说明index背后的list不是任何其他list的subset，就加入结果集。

时间O(n^3)

空间O(n)

Java实现

class Solution {
    public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
        List<Integer> res = new ArrayList<>();
        HashSet<Integer> subsetIndex = new HashSet<>();
        for (int i = 0; i < favoriteCompanies.size(); i++) {
            for (int j = 0; j < favoriteCompanies.size(); j++) {
                List<String> first = favoriteCompanies.get(i);
                List<String> second = favoriteCompanies.get(j);
                if (!first.equals(second)) {
                    if (first.size() < second.size()) {
                        if (isSubset(first, second)) {
                            subsetIndex.add(i);
                        }
                    }
                }
            }
        }
        for (int i = 0 ; i < favoriteCompanies.size(); i++) {
            if (!subsetIndex.contains(i)) {
                res.add(i);
            }
        }
        return res;
    }
    
    private boolean isSubset(List<String> listA, List<String> listB) {
        for (int i = 0; i < listA.size(); i++) {
            if (!listB.contains(listA.get(i))) {
                return false;
            }
        }
        return true;
    }
}

LeetCode 题目总结