<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 m[10001] = {}, valid[10001] = {}; // m <val, times>, valid <val, nextValidPos>
        for(int x : barcodes) ++m[x];
        for(int i=0, k=1; i<barcodes.size(); ++i){
            int c = 0, maxVal = 0;
            for(int ch=0; ch<10001; ++ch)
                if(maxVal<m[ch] && i>=valid[ch]) maxVal = m[ch], c = ch;
            if(!c) return {};
            --m[c];
            valid[c] = i+k+1;
            barcodes[i] = c;
        }
        return barcodes;
    }
};
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