<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> frequencySort(vector<int>& nums) {
    unordered_map<int, int> freq;

    // Count the frequency of each element
    for (int num : nums) {
        freq[num]++;
    }

    // Sort the array based on frequency and then by the element value
    sort(nums.begin(), nums.end(), [&](int a, int b) {
        if (freq[a] == freq[b])
            return a < b;  // If frequencies are the same, sort by the element itself
        return freq[a] > freq[b];  // Otherwise, sort by frequency (most frequent first)
    });

    return nums;
}
    vector<int> rearrangeBarcodes(vector<int>& barcodes) {
        int n=barcodes.size();
        vector<int> ans(n,0);
        barcodes=frequencySort(barcodes);

        for(int i=0;i<n;i=i+2)
        {
            ans[i]=barcodes[i/2];
        }
        for(int i=1;i<n;i=i+2)
        {
            if(n%2==0)
                ans[i]=barcodes[n/2+i/2];
            else 
                ans[i]=barcodes[n/2+i/2+1];
        }
        return ans;
    }
};
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