<p>Given a string <code>s</code>, rearrange the characters of <code>s</code> so that any two adjacent characters are not the same.</p>

<p>Return <em>any possible rearrangement of</em> <code>s</code> <em>or return</em> <code>&quot;&quot;</code> <em>if not possible</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "aab"
<strong>Output:</strong> "aba"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "aaab"
<strong>Output:</strong> ""
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 500</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    string reorganizeString(string s) {
        string ans=""; 
        map<char,int>map;
        for(int i=0;i<s.length();i++){
            map[s[i]]++;
        }

        vector< pair <int,char> > vect;
        for(auto i : map){
          vect.push_back(make_pair(i.second,i.first));
        }
 
    
         
        sort(vect.begin(),vect.end(),greater<pair <int,char>>());
        ans+=vect[0].second;
        if( vect[0].first>0){
             vect[0].first--;
        }
       for(int i=1;i<s.length();i++){
           sort(vect.begin(),vect.end(),greater<pair <int,char>>());
           if(ans[i-1]!=vect[0].second  ){
              ans.push_back(vect[0].second);
              vect[0].first--;
           }
           else{
               if(vect.size()==1  || vect[1].first<=0 ){
                   return "";
               }
               else{
                   ans.push_back(vect[1].second);
                   vect[1].first--;
              }
             
           }
            
       }

       return ans;
        
    }
};


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