<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() ;
        vector<int> ans( n , 0 ) ;

        if( n<=2 ) return barcodes ;

        map<int,int> mpp ;
        for( int i=0 ; i<n ; i++ ) mpp[barcodes[i]]++ ;

        int i = 0 ; 
        // int j = 1 ;

        priority_queue<pair<int,int>> pq ;
        for( auto & i : mpp ) pq.push( { i.second , i.first } ) ;

        // bool turn = 1 ;
        // while( !pq.empty() ){
        //     int n1 = pq.top().second ;
        //     int f1 = pq.top().first ;
        //     pq.pop() ;
        //     if( turn ){
        //         while( f1-- ){
        //             if( f1==1 ){ ans[i+1] = n1 ; }
        //             else{ ans[i] = n1 ; i=i+2 ; } 
        //         }
        //     }
        //     else{
        //         while( f1-- ){
        //             if( f1==1 ){ ans[j+1] = n1 ; }
        //             else{ ans[j] = n1 ; j=j+2 ; } 
        //         } 
        //     }
        //     cout<<i<<" "<<j<<endl ;
        //     turn = !turn ;
        //     // i=j ; j = j+2 ;
        //     // if( turn ){
        //     //     i=min(i,j) ; j=max(i,j) ;
        //     // }
        // }

        while( !pq.empty() ){
            int n1 = pq.top().second ;
            int f1 = pq.top().first ;
            pq.pop() ;  
            while( f1-- ){
                ans[i] = n1 ;
                i = i + 2 ;
                if( i>=ans.size() ) i = 1 ;
            }
            if( i>=ans.size() ) i = 1 ;
        }
        return ans ;
    }
};
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