<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) {
        vector<int> result(barcodes.size(),0);
        //Create hash map to count occurances
        unordered_map<int,int> mp;
        for(auto bc : barcodes){
            mp[bc] = mp[bc] + 1; //barcode number marks the index
        }
        //Create a set pair to get sorted count array
        set<pair<int,int>> st; 
        for(auto it = mp.begin() ; it != mp.end() ; it++){
            st.insert({it->second,it->first});
        }
        // Iterate from end to get sorted array in decreasing order
        int pos = 0;
        for(auto it = st.rbegin() ; it != st.rend() ; it++){
            //Add elements to list for number of available occurances
            for(int cnt = 0; cnt < it->first ; cnt++){
                if(pos >= barcodes.size())  // If size is reached, start filling odd loc
                    pos = 1;
                result[pos] = it->second;
                pos+=2; // Increment index by 2 so we fill even /odd locations
            }
        }
        return result;
    }
};
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