<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:
        s1 = list(s1)
        s2 = list(s2)
        queue = deque()
        queue.append([s1])
        steps = 0
        visited = set()
        visited.add("".join(s1))
        while queue:
            strings = queue.popleft()
            neighbors = []
            for string in strings:
                if string == s2:
                    return steps
                i = 0
                while i < len(string) and string[i] == s2[i]:
                    i += 1

                for j in range(i + 1, len(string)):
                    if string[j] == s2[i] and string[j] != s2[j]:
                        new_s = string[:]
                        new_s[i], new_s[j] = new_s[j], new_s[i]
                        if "".join(new_s) not in visited:
                            neighbors.append(new_s)
                            visited.add("".join(new_s))
                
            if neighbors:
                queue.append(neighbors)
            steps += 1
        
18300