<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>& bc) {
        
        int i=0;
        int curr=-1;

        int n = bc.size();

        while(i<n)
        {
            if(bc[i]==curr)
            {
                int j=i+1;
                while(j<n)
                {
                    if(bc[j]!=curr)
                    {
                        swap(bc[i], bc[j]);
                        break;
                    }
                    j++;
                }
            }
            curr = bc[i];
            i++;
        }

        reverse(bc.begin(), bc.end());
        i=0;
        curr=-1;

        while(i<n)
        {
            if(bc[i]==curr)
            {
                int j=i+1;
                while(j<n)
                {
                    if(bc[j]!=curr)
                    {
                        swap(bc[i], bc[j]);
                        break;
                    }
                    j++;
                }
            }
            curr = bc[i];
            i++;
        }

        reverse(bc.begin(), bc.end());

        return bc;

    }
};
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