<p>In a project, you have a list of required skills <code>req_skills</code>, and a list of people. The <code>i<sup>th</sup></code> person <code>people[i]</code> contains a list of skills that the person has.</p>

<p>Consider a sufficient team: a set of people such that for every required skill in <code>req_skills</code>, there is at least one person in the team who has that skill. We can represent these teams by the index of each person.</p>

<ul>
	<li>For example, <code>team = [0, 1, 3]</code> represents the people with skills <code>people[0]</code>, <code>people[1]</code>, and <code>people[3]</code>.</li>
</ul>

<p>Return <em>any sufficient team of the smallest possible size, represented by the index of each person</em>. You may return the answer in <strong>any order</strong>.</p>

<p>It is <strong>guaranteed</strong> an answer exists.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> req_skills = ["java","nodejs","reactjs"], people = [["java"],["nodejs"],["nodejs","reactjs"]]
<strong>Output:</strong> [0,2]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> req_skills = ["algorithms","math","java","reactjs","csharp","aws"], people = [["algorithms","math","java"],["algorithms","math","reactjs"],["java","csharp","aws"],["reactjs","csharp"],["csharp","math"],["aws","java"]]
<strong>Output:</strong> [1,2]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= req_skills.length &lt;= 16</code></li>
	<li><code>1 &lt;= req_skills[i].length &lt;= 16</code></li>
	<li><code>req_skills[i]</code> consists of lowercase English letters.</li>
	<li>All the strings of <code>req_skills</code> are <strong>unique</strong>.</li>
	<li><code>1 &lt;= people.length &lt;= 60</code></li>
	<li><code>0 &lt;= people[i].length &lt;= 16</code></li>
	<li><code>1 &lt;= people[i][j].length &lt;= 16</code></li>
	<li><code>people[i][j]</code> consists of lowercase English letters.</li>
	<li>All the strings of <code>people[i]</code> are <strong>unique</strong>.</li>
	<li>Every skill in <code>people[i]</code> is a skill in <code>req_skills</code>.</li>
	<li>It is guaranteed a sufficient team exists.</li>
</ul>

class Solution {
public:
vector<int>result;
unordered_map<string,int>dp;
void solve(vector<int>&people_skill,int destination,int idx,vector<int>&temp,int mask){
      if(idx==people_skill.size()){
         if(mask==destination){
            if(result.size()==0||result.size()>=temp.size()){
                result=temp;
            }
         }
         return;
      }
      string key=to_string(idx)+"_"+to_string(mask);
      if(dp.find(key)!=dp.end()){
        if(dp[key]<=temp.size()){
            return;
        }
      }
      if(result.size()!=0 && result.size()<=temp.size()){
        return;
      }
      //two option--take,not_take
      solve(people_skill,destination,idx+1,temp,mask);
      if((mask|people_skill[idx])!=mask){
      temp.push_back(idx);
      solve(people_skill,destination,idx+1,temp,mask|(people_skill[idx]));
      temp.pop_back();
      }
      if(temp.size()==0){
        dp[key]=-1;
      }
      else if(temp.size()!=1){
        dp[key]=temp.size();
      }
}
    vector<int> smallestSufficientTeam(vector<string>& req_skills, vector<vector<string>>& people) {
        //the team that we will select must have the reqd skills,as we are selecting the best things that we can do for this purpose is dp.
        //as the best thing possible to implement here is the concept of take and not take dp
        //we can easily solve this by brute-force dp,but bit-dp is the best implementation we can apply here for perfect optimisation.
        //we must return the least length of the array while fulfilling the conditions.
        int destination=pow(2,req_skills.size())-1; //111
        unordered_map<string,int>mpp;
        for(int i=0;i<req_skills.size();i++){
            mpp[req_skills[i]]=1<<i;
        }
        vector<int>people_skill(people.size());
        for(int i=0;i<people.size();i++){
            int skills=0;
            for(auto it:people[i]){
                skills=skills|mpp[it];
            }
            people_skill[i]=skills;
        }
        vector<int>temp;
        solve(people_skill,destination,0,temp,0);
        return result;
    }
};
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