<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:
        maxH = []
        for count, char in [(-a,'a'),(-b,'b'),(-c,'c')]:
            if count !=0 :
                heapq.heappush(maxH, (count,char))
        res = ''

        while maxH:
            cnt, char = heapq.heappop(maxH)
            if len(res)>1 and char == res[-1]==res[-2]:
                if not maxH:return res
                cnt2,char2= heapq.heappop(maxH)
                cnt2+=1
                res+=char2
                if cnt2!=0:
                    heapq.heappush(maxH,(cnt2,char2))
            else:
                res+=char
                cnt+=1
            if cnt!=0:
                heapq.heappush(maxH,(cnt,char))
        return res
16500