<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:
    /*
        word: idx1, idx2
        targetWord: <idx1, idx2>
        targetWord2: <idx3, idx4>
        1. hashmap<5*10^4*20> to create the inverted index
        2. 500*100 = 5*10^4*20 = (10^6)*(100) ~ 10^8
        3. intersection: 500 lists of 100 ints ~ 5*10^4
        google: 0,1,2,3
        fb: 0,2
        2
    */

    int isSubset(vector<string>&words, unordered_map<string,vector<int>>&invertedIndex)
    {
        vector<vector<int>>referenceList;
        int numLists = words.size();
        int numSuperSets = 0;
        for(auto &currWord: words)
        {
            referenceList.push_back(invertedIndex[currWord]);
        }
        unordered_map<int,int>referenceCountMap;
        for(auto &currReferenceList: referenceList)
        {
            for(auto reference: currReferenceList)
            {
                referenceCountMap[reference]++;
                if(referenceCountMap[reference]==numLists)
                {
                    numSuperSets++;
                }
            }
        }
        return numSuperSets;
    }

    vector<int> peopleIndexes(vector<vector<string>>& favoriteCompanies) {
        unordered_map<string,vector<int>>companyPersonMap;
        int n = favoriteCompanies.size();
        for(int i=0;i<n;i++)
        {
            for(auto &currCompany:favoriteCompanies[i])
            {
                companyPersonMap[currCompany].push_back(i);
            }
        }
        vector<int>resultList;
        for(int i = 0;i<n;i++)
        {
            if(isSubset(favoriteCompanies[i], companyPersonMap)==1)
            {
                resultList.push_back(i);
            }
        }
        return resultList;    
    }
};
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