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 1: Hash Table

We can map each company to a unique integer. Then, for each person, we convert their favorite companies into a set of integers. Finally, we check if the favorite companies of one person are a subset of another person's favorite companies.

The time complexity is $(n \times m \times k + n^2 \times m)$, and the space complexity is $O(n \times m)$. Here, $n$ and $m$ are the lengths of favoriteCompanies and the average length of each company's list, respectively, and $k$ is the average length of each company.

Python3JavaC++GoTypeScript

class Solution:
    def peopleIndexes(self, favoriteCompanies: List[List[str]]) -> List[int]:
        idx = 0
        d = {}
        n = len(favoriteCompanies)
        nums = [set() for _ in range(n)]
        for i, ss in enumerate(favoriteCompanies):
            for s in ss:
                if s not in d:
                    d[s] = idx
                    idx += 1
                nums[i].add(d[s])
        ans = []
        for i in range(n):
            if not any(i != j and (nums[i] & nums[j]) == nums[i] for j in range(n)):
                ans.append(i)
        return ans

class Solution {
    public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
        int n = favoriteCompanies.size();
        Map<String, Integer> d = new HashMap<>();
        int idx = 0;
        Set<Integer>[] nums = new Set[n];
        Arrays.setAll(nums, i -> new HashSet<>());
        for (int i = 0; i < n; ++i) {
            var ss = favoriteCompanies.get(i);
            for (var s : ss) {
                if (!d.containsKey(s)) {
                    d.put(s, idx++);
                }
                nums[i].add(d.get(s));
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            boolean ok = true;
            for (int j = 0; j < n && ok; ++j) {
                if (i != j && nums[j].containsAll(nums[i])) {
                    ok = false;
                }
            }
            if (ok) {
                ans.add(i);
            }
        }
        return ans;
    }
}

class Solution {
public:
    vector<int> peopleIndexes(vector<vector<string>>& favoriteCompanies) {
        int n = favoriteCompanies.size();
        unordered_map<string, int> d;
        int idx = 0;
        vector<unordered_set<int>> nums(n);

        for (int i = 0; i < n; ++i) {
            for (const auto& s : favoriteCompanies[i]) {
                if (!d.contains(s)) {
                    d[s] = idx++;
                }
                nums[i].insert(d[s]);
            }
        }

        auto check = [](const unordered_set<int>& a, const unordered_set<int>& b) {
            for (int x : a) {
                if (!b.contains(x)) {
                    return false;
                }
            }
            return true;
        };

        vector<int> ans;
        for (int i = 0; i < n; ++i) {
            bool ok = true;
            for (int j = 0; j < n && ok; ++j) {
                if (i != j && check(nums[i], nums[j])) {
                    ok = false;
                }
            }
            if (ok) {
                ans.push_back(i);
            }
        }

        return ans;
    }
};

func peopleIndexes(favoriteCompanies [][]string) (ans []int) {
    n := len(favoriteCompanies)
    d := make(map[string]int)
    idx := 0
    nums := make([]map[int]struct{}, n)

    for i := 0; i < n; i++ {
        nums[i] = make(map[int]struct{})
        for _, s := range favoriteCompanies[i] {
            if _, ok := d[s]; !ok {
                d[s] = idx
                idx++
            }
            nums[i][d[s]] = struct{}{}
        }
    }

    check := func(a, b map[int]struct{}) bool {
        for x := range a {
            if _, ok := b[x]; !ok {
                return false
            }
        }
        return true
    }
    for i := 0; i < n; i++ {
        ok := true
        for j := 0; j < n && ok; j++ {
            if i != j && check(nums[i], nums[j]) {
                ok = false
            }
        }
        if ok {
            ans = append(ans, i)
        }
    }

    return
}

function peopleIndexes(favoriteCompanies: string[][]): number[] {
    const n = favoriteCompanies.length;
    const d: Map<string, number> = new Map();
    let idx = 0;
    const nums: Set<number>[] = Array.from({ length: n }, () => new Set<number>());

    for (let i = 0; i < n; i++) {
        for (const s of favoriteCompanies[i]) {
            if (!d.has(s)) {
                d.set(s, idx++);
            }
            nums[i].add(d.get(s)!);
        }
    }

    const check = (a: Set<number>, b: Set<number>): boolean => {
        for (const x of a) {
            if (!b.has(x)) {
                return false;
            }
        }
        return true;
    };

    const ans: number[] = [];
    for (let i = 0; i < n; i++) {
        let ok = true;
        for (let j = 0; j < n && ok; j++) {
            if (i !== j && check(nums[i], nums[j])) {
                ok = false;
            }
        }
        if (ok) {
            ans.push(i);
        }
    }

    return ans;
}

Comments