<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>

#define ll long long
#define pb push_back
class Solution {
public:
    vector<int> rearrangeBarcodes(vector<int>& bar) {
        map<ll,ll>mp;
        ll n = bar.size();
        for(ll i=0;i<n;i++){
            mp[bar[i]]++;
        }
        vector<int>ans(n);
        vector<pair<ll,ll>>store;
        for(auto it:mp){
            store.pb({it.second,it.first});
        }
        sort(store.begin(),store.end());

        vector<ll>pokemon(n);
        ll index=0;
        for(ll i=0;i<store.size();i++){
            for(ll j=0;j<store[i].first;j++){
                pokemon[index]=store[i].second;
                index++;
            }
        }

        reverse(pokemon.begin(),pokemon.end());
        index = 0;
        for(ll i=0;i<n;i=i+2){
            ans[i]=pokemon[index];
            index++;
        }
        for(ll i=1;i<n;i=i+2){
            ans[i]=pokemon[index];
            index++;
        }
        return ans;

    }
};
50000