<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 {
public:
    vector<int> peopleIndexes(vector<vector<string>>& favoriteCompanies) {
        unordered_map<string, int> m;
        int id = 1, size = favoriteCompanies.size();

        // assign an integer id to each string company
        for (const auto& fc : favoriteCompanies) {
            for (const auto& c : fc) {
                if (m.count(c) > 0) {
                    continue;
                }
                m[c] = id++;
            }
        }

        // get bitset representation for each fav companies
        vector<bitset<50001>> vec;
        for (const auto& fc : favoriteCompanies) {
            bitset<50001> b;
            for (const auto& c : fc) {
                b[m[c]] = 1;
            }
            vec.push_back(b);        }

        vector<int> result;
        for (int i = 0; i < size; i++) {
            bool is_subset = false;
            for (int j = 0; j < size && !is_subset; j++) {
                if (i == j) {
                    continue;
                }
                // if ith is subset of jth, bitwise and will be same as ith (smaller one)
                if ((vec[i] & vec[j]) == vec[i]) {
                    is_subset = true;
                }
            }
            if (!is_subset) {
                result.push_back(i);
            }
        }
        return result;

    }
};
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