<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>& a) {
        map<int,int> mp;
        int n = a.size();
        for(int i =0 ;i<n;i++)
        {
            mp[a[i]]++;
        }
        set<pair<int,int>> s;
        for(auto it:mp)
        {
            s.insert({-1*it.second,it.first});
        }
        // for(auto it:s)
        // {
        //     cout<<it.first<<" "<<it.second<<endl;
        // }

        int i =0;
        int ct =0 ;
        vector<int> ans(n,0);
        while(s.size()>0)
        {
            auto it = s.begin();
            for(int j =0 ;j<-1*it->first;j++)
            {
                // cout<<i<<endl;
                if(i>=n)
                {
                    i = 1;
                }
                if(i<n)
                {
                    ct++;
                    ans[i]=it->second;
                }
                i+=2;
               
            }
            s.erase(it);
        }

        return ans;
        

    }
};
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