Distant Barcodes
<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>
Greedy; Array; Hash Table; Counting; Sorting; Heap (Priority Queue)
class Solution {
public:
    vector<int> rearrangeBarcodes(vector<int>& barcodes) {
        int mx = *max_element(barcodes.begin(), barcodes.end());
        int cnt[mx + 1];
        memset(cnt, 0, sizeof(cnt));
        for (int x : barcodes) {
            ++cnt[x];
        }
        sort(barcodes.begin(), barcodes.end(), [&](int a, int b) {
            return cnt[a] > cnt[b] || (cnt[a] == cnt[b] && a < b);
        });
        int n = barcodes.size();
        vector<int> ans(n);
        for (int k = 0, j = 0; k < 2; ++k) {
            for (int i = k; i < n; i += 2) {
                ans[i] = barcodes[j++];
            }
        }
        return ans;
    }
};