<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:
        def getNbr(s1,s2):
            idx = -1
            res = []
            for i , val in enumerate(s1):
                if s1[i] != s2[i]:
                    idx = i
                    break
            if idx != -1:
                for j in range(idx + 1, len(s1)):
                    if s1[j] == s2[idx]:
                        res.append(swap(s1,idx,j))
            
            return res
        
        def swap(s1,i,j):
            tmp = s1[i]
            s1 = s1[:i] + s1[j] + s1[i+1 :]
            s1 = s1[:j] + tmp + s1[j+1:]
            return s1



        qu = deque()
        level = 0
        vis = set()
        qu.append(s1)

        while qu:
            for _ in range(len(qu)):
                step = qu.popleft()
                if step in vis:
                    continue
                else:
                    vis.add(step)
                if step == s2:
                    return level
                
                for nbr in getNbr(step,s2):
                    qu.append(nbr)
            
            level += 1
        
        return -1
25500