<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 maxi = -1;
        int element = 0;

        unordered_map<int, int> freq;
        int n = barcodes.size();

        for (int i = 0; i < n; i++){
            freq[barcodes[i]]++;

            if (freq[barcodes[i]] > maxi){
                maxi = freq[barcodes[i]];
                element = barcodes[i];
            }
        }
        
        freq.erase(element);
        vector<int> res(n);
        int i;

        for (i = 0; i < 2 * maxi; i += 2){
            res[i] = element;
        }

        for (auto p : freq){
            for (int j = 0; j < p.second; j++){
                if (i >= n){
                    i = 1;
                }

                res[i] = p.first;
                i += 2;
            }
        }

        return res;
    }
};
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