<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:
        pairs = [['a', a], ['b', b], ['c', c]]
        pairs.sort(key=lambda x: -x[1])

        ans = ''
        previous = None 
        while True:
            index = 0
            
            letter, freq = pairs[index]
            minus = min(2, freq)

            while previous == letter:
                index += 1
                minus = 1
                letter, freq = pairs[index]
            previous = letter
            if freq <= 0:
                break

            ans += letter * minus
            pairs[index][1] -= minus
            pairs.sort(key=lambda x: -x[1])

        return ans
16600