<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) {
        unordered_map<int,int>mp;
        for(auto items : barcodes){
            mp[items]++;
        }
        vector<pair<int,int>>v(mp.begin(),mp.end());
        sort(v.begin(),v.end(),[](const pair<int,int> & a , const pair<int,int> &b){
            return a.second < b.second;
        });
        int j=v.size()-1;
        vector<int>ans(barcodes.size(),0);
        int i=0;
        while(i<barcodes.size() && j>=0){
            if(v[j].second>0){
                ans[i]=v[j].first;
                v[j].second--;
            }
            if(v[j].second==0) j--;
            i+=2;
        }
        i=1;
        while(i<barcodes.size() && j>=0){
            if(v[j].second>0){
                ans[i]=v[j].first;
                v[j].second--;
            }
            if(v[j].second==0) j--;
            i+=2;
        }
        return ans;
    }
};
45000