<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: //probably stay away from brick bbuildings. There's something better out there. 
    vector<int> rearrangeBarcodes(vector<int>& barcodes) {
    unordered_map<int, int> map;
    int prev = -1;
    vector<int> arranged;
    int val = -1;

    for(int i = 0; i < barcodes.size(); i++){
        map[barcodes[i]]++;}
    for(int i = 0; i < barcodes.size(); i++){
        auto it = map.begin();
        int max = 0;
        while(it != map.end()){
            if(max < it->second && prev != it->first){
                max = it->second;
                val = it->first;}
            it++;}
        prev = val;
        arranged.push_back(val);
        map[val]--;}
    return arranged;}};
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