K-Similar Strings
<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>
Breadth-First Search; Hash Table; String
func kSimilarity(s1 string, s2 string) int {
	next := func(s string) []string {
		i := 0
		res := []string{}
		for ; s[i] == s2[i]; i++ {
		}
		for j := i + 1; j < len(s1); j++ {
			if s[j] == s2[i] && s[j] != s2[j] {
				res = append(res, s[:i]+string(s[j])+s[i+1:j]+string(s[i])+s[j+1:])
			}
		}
		return res
	}

	q := []string{s1}
	vis := map[string]bool{s1: true}
	ans := 0
	for {
		for i := len(q); i > 0; i-- {
			s := q[0]
			q = q[1:]
			if s == s2 {
				return ans
			}
			for _, nxt := range next(s) {
				if !vis[nxt] {
					vis[nxt] = true
					q = append(q, nxt)
				}
			}
		}
		ans++
	}
}