<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, A: str, B: str) -> int:
        q = collections.deque()
        visited = set()
        q.append(A)
        visited.add(A)
        
        step = -1
        while q:
            step += 1
            lens = len(q)
            for _ in range(lens):
                curr = q.popleft()
                if curr == B:
                    return step
                for next in self._swap(curr, B):   # takes O(N)
                    if next in visited:
                        continue
                    q.append(next)
                    visited.add(next)
                
    def _swap(self, s, B):    # now only takes O(N)
        i = 0
        while s[i] == B[i]:   # if S[i]==B[i], we don't need to swap them - strong prune to makes sure swapped string always get more and more similar with B
            i += 1
        for j in range(i + 1, len(s)):
            if s[j] == B[i]:  # since B[i]!=s[i], if we swap s[j] to s[i], now B[i]=s[i]: this is how every swap make sure we get more and more closer to B
                yield s[:i] + s[j] + s[i+1:j] + s[i] + s[j+1:]
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