<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) {
      vector<int>hash(300,0);
        char ch='0';
        for(int i=0;i<s.size();i++){
            hash[s[i]-'a']++;
        }
        int maxi=*max_element(hash.begin(),hash.end());
            for(int i=0;i<hash.size();i++){
                if(hash[i]==maxi){
                     ch=i+'a';
                }
            }

        vector<char>l(s.size());
        int index=0;
        while(index<l.size() && maxi>0){
            l[index]=ch;
            index+=2;
            maxi--;
        }

        if(maxi>0){
            return "";
        }
        hash[ch-'a']=0;

         for(int i=0;i<300;i++){
            while(hash[i]>0){
                index=index>=s.size()?1:index;
                l[index]=i+'a';
                hash[i]--;
                index+=2;
            }
         }
         string d(l.begin(),l.end());
         return d;
    }
};
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