<p>In a warehouse, there is a row of barcodes, where the <code>i<sup>th</sup></code> barcode is <code>barcodes[i]</code>.</p>

<p>Rearrange the barcodes so that no two adjacent barcodes are equal. You may return any answer, and it is guaranteed an answer exists.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> barcodes = [1,1,1,2,2,2]
<strong>Output:</strong> [2,1,2,1,2,1]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> barcodes = [1,1,1,1,2,2,3,3]
<strong>Output:</strong> [1,3,1,3,1,2,1,2]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= barcodes.length &lt;= 10000</code></li>
	<li><code>1 &lt;= barcodes[i] &lt;= 10000</code></li>
</ul>

class Solution {
public:
    vector<int> rearrangeBarcodes(vector<int>& barcodes) {
      int n=barcodes.size();
      int maxi=INT_MAX;
      for(int i=0;i<n;i++){
        maxi=max(maxi,barcodes[i]);
    } 
  unordered_map<int,int>hsh;
      for(int i=0;i<n;i++)
      {
             hsh[barcodes[i]]++;
      }
      int prev=0;
      int cnt=0;
      int val=0;
      vector<int>op;
      for(int i=0;i<barcodes.size();i++)
      {
         auto it =hsh.begin();
         int mx=0;
         while(it!=hsh.end())
         {
            if(mx<it->second&&prev!=it->first)
            {
                mx=it->second;
                val=it->first;
            }
            it++;
         }
         
     prev=val;
op.push_back(val);
hsh[val]--;
      }
          return op; 
    }
};
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