不是任何其他给定列表的子集的列表计数

给定N个字符串列表，任务是找到不是任何其他给定列表的子列表的列表的计数。

例子：

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.

方法
请按照以下步骤解决问题：

1. 首先，枚举所有向量中所有可能的字符串，即给它们分配一个整数。
2. 然后，对所有单独的列表使用一个 Bitset 来存储它们中存在的字符串。
3. 比较位集。如果其中一个位集是另一个的子集，则忽略该列表。否则，在集合中插入该列表的索引。
4. 打印集合中的所有索引。

下面的代码是上述方法的实现：

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& 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 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;
}

输出：

0 1

时间复杂度： O ( N * M )
辅助空间： O ( N * M )