<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>& arr) {
        if(arr.size()==1)return arr;
       map<int,int>mp;
       for(int i=0;i<arr.size();i++){
        mp[arr[i]]++;
       } 
       vector<pair<int,int>>vec;
       for(auto &it:mp){
        vec.push_back({it.second,it.first});
       }
       sort(vec.begin(),vec.end());
       reverse(vec.begin(),vec.end());
       int n=arr.size();
       vector<int>ans(n,0);
       int i=0;
       int x=vec[0].first;
       int y=vec[1].first;
       int z=0;
       if(vec.size()>2){
         z=vec[2].first;
       }
       
       int cnt=0;
       while(i<n){
        for(int j=0;j<vec.size();j++){
            if(j==1)cnt++;
            if(j!=2&&vec[j].first>0){
                ans[i++]=vec[j].second;
                vec[j].first--;
            }
            if(j==2&&(cnt<=z-x+y||cnt>y)){
                if(vec[j].first>0){
                  ans[i++]=vec[j].second;
                vec[j].first--;
                }
               
            }
            
        }
       }
       return ans;
    }
};
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