<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>

bool func(unordered_map<int,int>& amap1,unordered_map<int,int>& amap2){
    return amap1.size()<amap2.size();
}
class Solution {
public:
    vector<int> peopleIndexes(vector<vector<string>>& favoriteCompanies) {
        map<string,int>amap;
        int count=1;
        int n=favoriteCompanies.size();
        for(int i=0;i<favoriteCompanies.size();i++){
            for(int j=0;j<favoriteCompanies[i].size();j++){
                amap[favoriteCompanies[i][j]]++;
            }
        }
        for(auto i:amap){
            amap[i.first]=count;
            count++;
        }
        vector<unordered_map<int,int>>fav;
        for(int i=0;i<n;i++){
            unordered_map<int,int> amap1;
            for(int j=0;j<favoriteCompanies[i].size();j++){
                amap1[amap[favoriteCompanies[i][j]]]=1;
            }
            fav.push_back(amap1);
        }
        // sort(fav.begin(),fav.end(),func);
        vector<int>ans;

        for(int i=0;i<fav.size();i++){
            int check=0;
            for(int j=0;j<fav.size();j++){
                if(i==j){
                    continue;
                }
                else{
                    int flag=0;
                    for(auto k:fav[i]){
                        if(fav[j].count(k.first)==0){
                            flag=1;
                            break;
                        }
                    }
                    if(flag==0){
                        check=1;
                        break;
                    }
                }
            }
            if(check==0){
                ans.push_back(i);
            }
        }

        return ans;
    }
};
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