<p>A string <code>s</code> is called <strong>happy</strong> if it satisfies the following conditions:</p>

<ul>
	<li><code>s</code> only contains the letters <code>&#39;a&#39;</code>, <code>&#39;b&#39;</code>, and <code>&#39;c&#39;</code>.</li>
	<li><code>s</code> does not contain any of <code>&quot;aaa&quot;</code>, <code>&quot;bbb&quot;</code>, or <code>&quot;ccc&quot;</code> as a substring.</li>
	<li><code>s</code> contains <strong>at most</strong> <code>a</code> occurrences of the letter <code>&#39;a&#39;</code>.</li>
	<li><code>s</code> contains <strong>at most</strong> <code>b</code> occurrences of the letter <code>&#39;b&#39;</code>.</li>
	<li><code>s</code> contains <strong>at most</strong> <code>c</code> occurrences of the letter <code>&#39;c&#39;</code>.</li>
</ul>

<p>Given three integers <code>a</code>, <code>b</code>, and <code>c</code>, return <em>the <strong>longest possible happy </strong>string</em>. If there are multiple longest happy strings, return <em>any of them</em>. If there is no such string, return <em>the empty string </em><code>&quot;&quot;</code>.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within a string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> a = 1, b = 1, c = 7
<strong>Output:</strong> &quot;ccaccbcc&quot;
<strong>Explanation:</strong> &quot;ccbccacc&quot; would also be a correct answer.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> a = 7, b = 1, c = 0
<strong>Output:</strong> &quot;aabaa&quot;
<strong>Explanation:</strong> It is the only correct answer in this case.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= a, b, c &lt;= 100</code></li>
	<li><code>a + b + c &gt; 0</code></li>
</ul>

class Solution:
    def longestDiverseString(self, a: int, b: int, c: int) -> str:
        res = []
        heap = []
        for count, char  in [(-a,"a"), (-b,"b"), (-c,"c")]:
            if count != 0:
                heapq.heappush(heap,(count,char))
            
        while heap:
            count, char = heapq.heappop(heap)

            if len(res) > 1 and res[-2] == res[-1] == char:
                if not heap:
                    break
                count2,char2 = heapq.heappop(heap)
                res.append(char2)
                count2+=1
                if count2< 0:
                    heapq.heappush(heap,(count2,char2))
                heapq.heappush(heap,(count,char))

            else:
                count+=1 
                res.append(char)
                if count < 0:
                    heapq.heappush(heap,(count,char))

        return "".join(res)
16600