<p>Given the array <code>favoriteCompanies</code> where <code>favoriteCompanies[i]</code> is the list of favorites companies for the <code>ith</code> person (<strong>indexed from 0</strong>).</p>

<p><em>Return the indices of people whose list of favorite companies is not a <strong>subset</strong> of any other list of favorites companies</em>. You must return the indices in increasing order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> favoriteCompanies = [[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;],[&quot;google&quot;,&quot;microsoft&quot;],[&quot;google&quot;,&quot;facebook&quot;],[&quot;google&quot;],[&quot;amazon&quot;]]
<strong>Output:</strong> [0,1,4] 
<strong>Explanation:</strong> 
Person with index=2 has favoriteCompanies[2]=[&quot;google&quot;,&quot;facebook&quot;] which is a subset of favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] corresponding to the person with index 0. 
Person with index=3 has favoriteCompanies[3]=[&quot;google&quot;] which is a subset of favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;] and favoriteCompanies[1]=[&quot;google&quot;,&quot;microsoft&quot;]. 
Other lists of favorite companies are not a subset of another list, therefore, the answer is [0,1,4].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> favoriteCompanies = [[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;],[&quot;leetcode&quot;,&quot;amazon&quot;],[&quot;facebook&quot;,&quot;google&quot;]]
<strong>Output:</strong> [0,1] 
<strong>Explanation:</strong> In this case favoriteCompanies[2]=[&quot;facebook&quot;,&quot;google&quot;] is a subset of favoriteCompanies[0]=[&quot;leetcode&quot;,&quot;google&quot;,&quot;facebook&quot;], therefore, the answer is [0,1].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> favoriteCompanies = [[&quot;leetcode&quot;],[&quot;google&quot;],[&quot;facebook&quot;],[&quot;amazon&quot;]]
<strong>Output:</strong> [0,1,2,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= favoriteCompanies.length &lt;= 100</code></li>
	<li><code>1 &lt;= favoriteCompanies[i].length &lt;= 500</code></li>
	<li><code>1 &lt;= favoriteCompanies[i][j].length &lt;= 20</code></li>
	<li>All strings in <code>favoriteCompanies[i]</code> are <strong>distinct</strong>.</li>
	<li>All lists of favorite companies are <strong>distinct</strong>, that is, If we sort alphabetically each list then <code>favoriteCompanies[i] != favoriteCompanies[j].</code></li>
	<li>All strings consist of lowercase English letters only.</li>
</ul>

class Solution {
    struct record_item {
        int idx, nC;
        int *bits;
    };

public:
    vector<int> peopleIndexes(vector<vector<string>> &favoriteCompanies)
    {
        vector<int> *v;
        struct record_item *w;
        unordered_map<string, int> *t;
        int b, c, d, i, j, k, n, s;
        
        n = favoriteCompanies.size();
        t = new unordered_map<string, int>();
        for (i = 0; i < n; i++)
            for (j = 0; j < favoriteCompanies[i].size(); j++)
                if (t->find(favoriteCompanies[i][j]) == t->end())
                    t->insert({ favoriteCompanies[i][j], t->size() });
        s = (t->size() + 29) / 30;
        w = new record_item[n];
        for (i = 0; i < n; i++) {
            w[i].idx = i;
            w[i].nC = favoriteCompanies[i].size();
            w[i].bits = new int[s];
            std::fill(&w[i].bits[0], &w[i].bits[s], 0);
            for (j = 0; j < w[i].nC; j++) {
                c = (*t)[favoriteCompanies[i][j]];
                w[i].bits[c / 30] |= 1 << (c % 30);
            }
        }
        std::sort(&w[0], &w[n],
                  [](struct record_item a, struct record_item b) {
                      return a.nC < b.nC;
                  });

        v = new vector<int>();
        for (i = 0; i < n - 1; i++) {
            for (j = i + 1; j < n; j++) {
                for (k = 0; k < s; k++)
                    if ((w[i].bits[k] & w[j].bits[k]) != w[i].bits[k])
                        break;
                if (k == s)
                    break;
            }
            if (j == n)
                v->push_back(w[i].idx);
            delete[] w[i].bits;
        }
        v->push_back(w[n - 1].idx);
        std::sort(v->begin(), v->end());
        delete[] w[n - 1].bits;
        delete[] w;
        delete t;

        return *v;
    }
};
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