<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
        
        src = ''
        tar = ''
        for i in range(len(s1)):
            if s1[i] != s2[i]:
                src += s1[i]
                tar += s2[i]
        
        q = [(src, tar, 0)]
        visited = {(src, tar)}

        while q:
            # print(q)
            src, tar, cnt = q.pop(0)

            for newSrc, newTar in self.getNxt(src, tar):
                if not newSrc:
                    return cnt + 1
                if (newSrc, newTar) not in visited:
                    visited.add((newSrc, newTar))
                    q.append((newSrc, newTar, cnt+1))
        
        return -1
    
    def getNxt(self, src, tar):
        for i in range(1, len(src)-1):
            if src[0] == tar[i]:
                if src[i] == tar[0]:
                    yield (src[1:i]+src[i+1:], tar[1:i]+tar[i+1:])
                else:
                    yield(src[i] + src[1:i] + src[i+1:], tar[:i] + tar[i+1:])
        ptr = len(src)-1
        if src[0] == tar[ptr]:
            if src[ptr] == tar[0]:
                yield (src[1:ptr], tar[1:ptr])
            else:
                yield(src[ptr] + src[1:ptr], tar[:ptr])
                    

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