<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 pos=0;
        map<int,int>mpp;
        for(auto e:barcodes)mpp[e]++;
        set<pair<int,int>>st;
        for(auto e:mpp){
            st.insert({e.second,e.first});
        }
        for(auto it=st.rbegin();it!=st.rend();++it){
            //this ensures that first fill even position and then odd position
            for(auto cnt=0;cnt<it->first;cnt++,pos+=2){
                if(pos>=barcodes.size()){
                    pos=1;
                }
                barcodes[pos]=it->second;
            }
        }
        return barcodes;
    }
};
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