<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) {
         
    int hash[26]={0};
     for(int i =0;i<s.length();i++){
          hash[s[i] -'a']++;}
    char  mostrepeatchar;
    int  freq = INT_MIN;
    
       for(int i=0;i<26;i++){
        if(hash[i]>freq){
            freq=hash[i];
           mostrepeatchar =i+'a';
        }
       }
       int index=0;
       while(freq>0&&index<s.size()){
        s[index]=mostrepeatchar;
        freq--;
        index+=2;
       }
       if(freq!=0){
        return "";
       }

       hash[mostrepeatchar-'a']=0;
       for(int i=0;i<26;i++){
        while(hash[i]>0){
            index= index>=s.length() ? 1:index;
                s[index]=i +'a';
                hash[i]--;
                index+=2;
                 
            
        }
       }
       
        return s;

    }
};
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