<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:

    bool isSubset(vector<int> num1,vector<int> num2){
        int n1 = num1.size(); int n2 = num2.size();
        int p1 = 0; int p2 = 0;
        while(p1<n1 && p2<n2){
            if(num1[p1]==num2[p2]){
                p1++;
                p2++;
            }
            else if(num1[p1]<num2[p2]){
                p1++;
            }
            else{
                return false;
            }
        }
       if(p2==n2){
        return true;
       }
       return false;
    }

    vector<int> peopleIndexes(vector<vector<string>>& fc) {
        int n = fc.size();
        map<string,int> mp;
        int num = 0;
        for(int i=0;i<n;i++){
            int m = fc[i].size();
            for(int j =0;j<m;j++){
                if(mp.find(fc[i][j])==mp.end()){
                    mp[fc[i][j]] = num;
                    num++;
                }
            }
        }

        vector<vector<int>> v;
        for(int i=0;i<n;i++){
            int m = fc[i].size();
            vector<int> temp;
            for(int j =0;j<m;j++){
                temp.push_back(mp[fc[i][j]]);
            }
            sort(temp.begin(),temp.end());
            v.push_back(temp);
        }
        vector<int> ans;
        for(int i=0;i<n;i++){
            bool contains = false;
            for(int j=0;j<n;j++){
                if(i!=j && isSubset(v[j],v[i])){
                    contains = true;
                    break;
                }
            }
            if(contains==false){
                ans.push_back(i);
            }
        }
        return ans;

        
    }
};
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