<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>freq(10001,0);
		vector<pair<int,int>>v;
		for(int i=0;i<n;i++){
			freq[barcodes[i]]++;
		}
		for(int i=10000;i>=0;i--){
			if(freq[i]==0)
				continue;
			v.push_back({freq[i],i});
		}
		sort(v.rbegin(),v.rend());
		int m = v.size();
		bool flag = false;
		int ind = 0;
		for(int i=0;i<m;i++){
			while(v[i].first){
				barcodes[ind] = v[i].second;
				if(ind+2 >= n)
				ind = 1;
				else
				ind += 2;
			    v[i].first--;
			}
		}
		return barcodes;
    }
};
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