<p>You are given three integers <code>x</code>, <code>y</code>, and <code>z</code>.</p>

<p>You have <code>x</code> strings equal to <code>&quot;AA&quot;</code>, <code>y</code> strings equal to <code>&quot;BB&quot;</code>, and <code>z</code> strings equal to <code>&quot;AB&quot;</code>. You want to choose some (possibly all or none) of these strings and concatenate them in some order to form a new string. This new string must not contain <code>&quot;AAA&quot;</code> or <code>&quot;BBB&quot;</code> as a substring.</p>

<p>Return <em>the maximum possible length of the new string</em>.</p>

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

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

<pre>
<strong>Input:</strong> x = 2, y = 5, z = 1
<strong>Output:</strong> 12
<strong>Explanation: </strong>We can concactenate the strings &quot;BB&quot;, &quot;AA&quot;, &quot;BB&quot;, &quot;AA&quot;, &quot;BB&quot;, and &quot;AB&quot; in that order. Then, our new string is &quot;BBAABBAABBAB&quot;. 
That string has length 12, and we can show that it is impossible to construct a string of longer length.
</pre>

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

<pre>
<strong>Input:</strong> x = 3, y = 2, z = 2
<strong>Output:</strong> 14
<strong>Explanation:</strong> We can concactenate the strings &quot;AB&quot;, &quot;AB&quot;, &quot;AA&quot;, &quot;BB&quot;, &quot;AA&quot;, &quot;BB&quot;, and &quot;AA&quot; in that order. Then, our new string is &quot;ABABAABBAABBAA&quot;. 
That string has length 14, and we can show that it is impossible to construct a string of longer length.
</pre>

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

<ul>
	<li><code>1 &lt;= x, y, z &lt;= 50</code></li>
</ul>

class Solution:
    def longestString(self, x: int, y: int, z: int) -> int:
        
        @cache
        def dfs(a, b, c, p):
            if 0 == a == b == c:
                return 0

            i = j = 0
            if p != "a":
                if a:
                    i = dfs(a - 1, b, c, "a") + 2
                if c:
                    j = dfs(a, b, c - 1, "b") + 2
            if b and p != "b":
                i = dfs(a, b - 1, c, "b") + 2

            return max(i, j)

        return dfs(x, y, z, "")
        


1414