<p>Given two integers <code>a</code> and <code>b</code>, return <strong>any</strong> string <code>s</code> such that:</p>

<ul>
	<li><code>s</code> has length <code>a + b</code> and contains exactly <code>a</code> <code>&#39;a&#39;</code> letters, and exactly <code>b</code> <code>&#39;b&#39;</code> letters,</li>
	<li>The substring <code>&#39;aaa&#39;</code> does not occur in <code>s</code>, and</li>
	<li>The substring <code>&#39;bbb&#39;</code> does not occur in <code>s</code>.</li>
</ul>

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

<pre>
<strong>Input:</strong> a = 1, b = 2
<strong>Output:</strong> &quot;abb&quot;
<strong>Explanation:</strong> &quot;abb&quot;, &quot;bab&quot; and &quot;bba&quot; are all correct answers.
</pre>

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

<pre>
<strong>Input:</strong> a = 4, b = 1
<strong>Output:</strong> &quot;aabaa&quot;
</pre>

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

<ul>
	<li><code>0 &lt;= a, b &lt;= 100</code></li>
	<li>It is guaranteed such an <code>s</code> exists for the given <code>a</code> and <code>b</code>.</li>
</ul>

class Solution:
    def strWithout3a3b(self, a: int, b: int) -> str:
        if a < 3 and b < 3:
            return a * 'a' + b * 'b'
        
        ans = []
        while a or b:
            if len(ans) >= 2 and ans[-1] == ans[-2]:
                if ans[-1] == 'a':
                    ans.append('b')
                    b -= 1
                else:
                    ans.append('a')
                    a -= 1
            elif a > b:
                ans.append('a')
                a -= 1
            elif a <= b:
                ans.append('b')
                b -= 1
        
        return ''.join(ans)
16300