<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:
    ans = 0
    q = collections.deque([s1])
    seen = {s1}

    while q:
      for _ in range(len(q)):
        curr = q.popleft()
        if curr == s2:
          return ans
        for child in self._getChildren(curr, s2):
          if child in seen:
            continue
          q.append(child)
          seen.add(child)
      ans += 1

    return -1

  def _getChildren(self, curr: str, target: str) -> List[str]:
    children = []
    s = list(curr)
    i = 0  # the first index s.t. curr[i] != target[i]
    while curr[i] == target[i]:
      i += 1

    for j in range(i + 1, len(s)):
      if s[j] == target[i]:
        s[i], s[j] = s[j], s[i]
        children.append(''.join(s))
        s[i], s[j] = s[j], s[i]

    return children
        
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