<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();
        map<int,int> mp;
        for(int i=0;i<n;i++){
            mp[barcodes[i]]++;
        }
        int maxidx=mp.begin()->first;
        for(auto x:mp){
            if(mp[maxidx]<x.second){
                maxidx=x.first;
            }
        }
        int ct=0;
        for(auto x:mp){
            if(x.first!=maxidx)ct+=x.second;
        }
        vector<int> ans;
        while(mp[maxidx]--){
            ans.push_back(maxidx);
            for(auto &x:mp){
                if(x.first!=maxidx&&ct>=mp[maxidx]){
                    x.second--;
                    ans.push_back(x.first);
                    if(x.second==0)mp.erase(x.first);
                    ct--;
                }
            }
        }
        return ans;
    }
};
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