<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>& s) {
        vector<int> freq(10001,0);

        for(int i=0;i<s.size();i++)
            freq[s[i]]++;
        
        int maxi = 0;
        int letter=0;
        for(int i=0;i<10001;i++) {
            if(freq[i] > maxi) {
                maxi = freq[i];
                letter = i;
            }
        }

        vector<int> ans(s.size(), 0);
        int idx=0;

        while(freq[letter] > 0) {
            ans[idx] = letter;
            idx += 2;
            freq[letter]--;
        }

        for(int i=0;i<10001;i++) {
            while(freq[i]--) {
                if(idx >= s.size())
                    idx = 1;
                ans[idx] = i;
                idx += 2;
            }
        }

        return ans;       
    }
    
};
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