<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>& a) {
        int n = a.size();
        map<int, int,greater<int>> mp;
        int mx = 0, in = 0;
        for (int i = 0; i < n; ++i) {
            mp[a[i]]++;
            if (mp[a[i]] > mx) {
                mx = mp[a[i]];
                in = a[i];
            }
        }

        mp[in] = 0;
        vector <int> ans;
        int d = (n - mx) / mx;
        int rem = (n - mx) % mx;
        vector<int> res;
        while(1)
        {
            bool flag=0;
            for(auto x:mp)
            {
                if(x.second>0)
                {
                    res.push_back(x.first);
                    mp[x.first]--;
                    flag=1;
                }
            }
            if(!flag)break;
        }

        int j=0;
        for (int i = 0; i < mx; ++i) {
            ans.push_back(in);
            int dd = d;
            if(rem>0)
            {
                dd++;
                rem--;
            }
            while (dd > 0) {
                ans.push_back(res[j++]);
                dd--;
            }
        }

        return ans;
    }
};
48900