# Count of lists which are not a subset of any other given lists

Given N lists of strings, the task is to find the count of lists which are not a sublist of any other given lists.

Examples:

Input: [[“hey”, “hi”, “hello”], [“hey”, “bye”], [“hey”, “hi”]]
Output: 2
Explaination
The third list is a subset of the first list, hence the first and the second list are the required lists.

Input: [[“geeksforgeeks”, “geeks”], [“geeks”, “geeksforgeeks”]]
Output: 0
Explanation: Both the lists comprise of same set of strings.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach
Follow the steps below to solve the problem:

1. First of all, enumerate all the possible strings in all vectors, i.e. assign them an integer.
2. Then, use a Bitset for all individual lists to store the strings present in them.
3. Compare the bitsets. If one of the bitsets is a subset of another, ignore that list. Otherwise insert the index of that list in a set.
4. Print all the indices in the set.

Below code is the implementation of the above approach:

## C++

 `// C++ program to find all lists ` `// which are not a subset of any ` `// other given lists ` `#include ` `using` `namespace` `std; ` ` `  `#define N 50005 ` ` `  `// Function to print all lists which ` `// are not a subset of any other lists ` `void` `findNonSubsets(vector >& v, ` `                    ``vector<``int``>& ans) ` `{ ` `    ``unordered_map mp; ` `    ``int` `id = 1; ` `    ``// Enumerate all strings ` `    ``// present in all lists ` `    ``for` `(``int` `i = 0; i < v.size(); i++) { ` `        ``for` `(``int` `j = 0; j < v[i].size(); j++) { ` `            ``if` `(mp.count(v[i][j]) > 0) ` `                ``continue``; ` ` `  `            ``mp[v[i][j]] = id++; ` `        ``} ` `    ``} ` ` `  `    ``// Compute and store bitsets ` `    ``// of all strings in lists ` `    ``vector > v1; ` ` `  `    ``for` `(``int` `i = 0; i < v.size(); i++) { ` `        ``bitset b; ` `        ``for` `(``int` `j = 0; j < v[i].size(); j++) { ` `            ``b[mp[v[i][j]]] = 1; ` `        ``} ` `        ``v1.push_back(b); ` `    ``} ` `    ``for` `(``int` `i = 0; i < v.size(); i++) { ` `        ``bool` `flag = ``false``; ` `        ``for` `(``int` `j = 0; !flag and j < v.size(); j++) { ` `            ``if` `(i != j) { ` `                ``// If one of the bitsets is ` `                ``// a subset of another, the ` `                ``// logical AND is equal to the ` `                ``// subset(intersection operation) ` `                ``if` `((v1[i] & v1[j]) == v1[i]) { ` `                    ``flag = ``true``; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``if` `(!flag) { ` `            ``ans.push_back(i); ` `        ``} ` `    ``} ` `    ``return``; ` `} ` ` `  `// Driver Program ` `signed` `main() ` `{ ` `    ``vector > v ` `        ``= { { ``"hey"``, ``"hello"``, ``"hi"` `}, ` `            ``{ ``"hey"``, ``"bye"` `}, ` `            ``{ ``"hey"``, ``"hi"` `} }; ` ` `  `    ``vector<``int``> ans; ` `    ``findNonSubsets(v, ans); ` ` `  `    ``if` `(ans.size() == 0) { ` `        ``cout << -1 << endl; ` `        ``return` `0; ` `    ``} ` ` `  `    ``for` `(``int` `i = 0; i < ans.size(); i++) { ` `        ``cout << ans[i] << ``" "``; ` `    ``} ` ` `  `    ``return` `0; ` `} `

Output:

```0 1
```

Time Complexity: O ( N * M )
Auxiliary Space: O ( N * M ) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : nidhi_biet