<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 {
 public:
  int kSimilarity(string s1, string s2) {
    int ans = 0;
    queue<string> q{{s1}};
    unordered_set<string> seen{{s1}};

    while (!q.empty()) {
      for (int sz = q.size(); sz > 0; --sz) {
        string curr = q.front();
        q.pop();
        if (curr == s2)
          return ans;
        for (const string& child : getChildren(curr, s2)) {
          if (seen.contains(child))
            continue;
          q.push(child);
          seen.insert(child);
        }
      }
      ++ans;
    }

    return -1;
  }

 private:
  vector<string> getChildren(string& curr, const string& target) {
    vector<string> children;
    int i = 0;  // the first index s.t. curr[i] != target[i]
    while (curr[i] == target[i])
      ++i;

    for (int j = i + 1; j < curr.length(); ++j)
      if (curr[j] == target[i]) {
        swap(curr[i], curr[j]);
        children.push_back(curr);
        swap(curr[i], curr[j]);
      }

    return children;
  }
};
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