Longest Happy String
<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>
Greedy; String; Heap (Priority Queue)
class Solution:
    def longestDiverseString(self, a: int, b: int, c: int) -> str:
        h = []
        if a > 0:
            heappush(h, [-a, 'a'])
        if b > 0:
            heappush(h, [-b, 'b'])
        if c > 0:
            heappush(h, [-c, 'c'])

        ans = []
        while len(h) > 0:
            cur = heappop(h)
            if len(ans) >= 2 and ans[-1] == cur[1] and ans[-2] == cur[1]:
                if len(h) == 0:
                    break
                nxt = heappop(h)
                ans.append(nxt[1])
                if -nxt[0] > 1:
                    nxt[0] += 1
                    heappush(h, nxt)
                heappush(h, cur)
            else:
                ans.append(cur[1])
                if -cur[0] > 1:
                    cur[0] += 1
                    heappush(h, cur)

        return ''.join(ans)
