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
class Solution {
public:
    int kSimilarity(string s1, string s2) {
        queue<string> q{{s1}};
        unordered_set<string> vis{{s1}};
        int ans = 0;
        while (1) {
            for (int i = q.size(); i; --i) {
                auto s = q.front();
                q.pop();
                if (s == s2) {
                    return ans;
                }
                for (auto& nxt : next(s, s2)) {
                    if (!vis.count(nxt)) {
                        vis.insert(nxt);
                        q.push(nxt);
                    }
                }
            }
            ++ans;
        }
    }

    vector<string> next(string& s, string& s2) {
        int i = 0, n = s.size();
        for (; s[i] == s2[i]; ++i) {}
        vector<string> res;
        for (int j = i + 1; j < n; ++j) {
            if (s[j] == s2[i] && s[j] != s2[j]) {
                swap(s[i], s[j]);
                res.push_back(s);
                swap(s[i], s[j]);
            }
        }
        return res;
    }
};