<p>Given a string <code>s</code>, rearrange the characters of <code>s</code> so that any two adjacent characters are not the same.</p>

<p>Return <em>any possible rearrangement of</em> <code>s</code> <em>or return</em> <code>&quot;&quot;</code> <em>if not possible</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "aab"
<strong>Output:</strong> "aba"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "aaab"
<strong>Output:</strong> ""
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 500</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def reorganizeString(self, s: str) -> str:
        # hash_map = [0 for i in range(26)]
        # n = len(s)
        # for i in range(n):
        #     hash_map[ord(s[i]) - ord("a")] += 1
        ans = []
        pq = [(-count, char) for char, count in Counter(s).items()]
        heapify(pq)

        while pq:
            count_first, char_first = heappop(pq)
            if not ans or char_first != ans[-1]:
                ans.append(char_first)
                if count_first + 1 != 0:
                    heappush(pq, (count_first + 1, char_first))
            else:
                if not pq: return ''
                count_second, char_second = heappop(pq)
                ans.append(char_second)
                if count_second + 1 != 0:
                    heappush(pq, (count_second + 1, char_second))
                heappush(pq, (count_first,char_first))
        
        return ''.join(ans)
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