<p>Strings <code>s1</code> and <code>s2</code> are <code>k</code><strong>-similar</strong> (for some non-negative integer <code>k</code>) if we can swap the positions of two letters in <code>s1</code> exactly <code>k</code> times so that the resulting string equals <code>s2</code>.</p>

<p>Given two anagrams <code>s1</code> and <code>s2</code>, return the smallest <code>k</code> for which <code>s1</code> and <code>s2</code> are <code>k</code><strong>-similar</strong>.</p>

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

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;ba&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> The two string are 1-similar because we can use one swap to change s1 to s2: &quot;ab&quot; --&gt; &quot;ba&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;abc&quot;, s2 = &quot;bca&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The two strings are 2-similar because we can use two swaps to change s1 to s2: &quot;abc&quot; --&gt; &quot;bac&quot; --&gt; &quot;bca&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s1.length &lt;= 20</code></li>
	<li><code>s2.length == s1.length</code></li>
	<li><code>s1</code> and <code>s2</code> contain only lowercase letters from the set <code>{&#39;a&#39;, &#39;b&#39;, &#39;c&#39;, &#39;d&#39;, &#39;e&#39;, &#39;f&#39;}</code>.</li>
	<li><code>s2</code> is an anagram of <code>s1</code>.</li>
</ul>

class Solution:
    def kSimilarity(self, s1: str, s2: str) -> int:
        if s1 == s2:
            return 0

        keep = tuple(ch1 != ch2 for ch1, ch2 in zip(s1, s2))
        s = sum((ord(ch) - ord('a')) << (3*i) for i, ch
                in enumerate(compress(s1, keep)))
        t = sum((ord(ch) - ord('a')) << (3*i) for i, ch
                in enumerate(compress(s2, keep)))

        i = -3
        lim = sum(keep)*3

        @cache
        def kSim(s: int):
            nonlocal i
            i += 3
            if s == t:
                r = 0
            elif s & 7 << i == t & 7 << i:
                r = kSim(s)
            else:
                r = 1 + min(kSim((s & 7 << j) >> (j-i) |
                                 (s & 7 << i) << (j-i) |
                                 s & ~(7 << i | 7 << j))
                            for j in range(i + 3, lim, 3) if
                            (s & 7 << j != t & 7 << j and
                            (s & 7 << j) >> (j-i) == t & 7 << i))
            i -= 3
            return r

        return kSim(s)

18800