<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) {

        map<int, int> M;
        int max_ele, count_max = 0;
        for(const auto& x : barcodes) 
        {
            M[x]++;
            if(M[x] > count_max) {
                count_max = M[x];
                max_ele = x;
            }
        }

        int ind = 0;
        int n = barcodes.size();
        for(int i = 0; i < count_max; i++) {
            barcodes[ind] = max_ele;
            if(ind + 2 >= n) ind = 1;
            else ind += 2;
        }
        for(auto itr = M.begin(); itr != M.end(); ++itr) {
            if(itr->first == max_ele)continue;
            for(int i = 0; i < itr->second; i++) {
                barcodes[ind] = itr->first;
                if(ind + 2 >= n) ind = 1;
                else ind += 2;
            }
        }

        return barcodes;
        
    }
};
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