<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:
    unordered_set<string> getNext(string& s1, string& s2) {
        unordered_set<string> res;
        for (int i = 0; i < s1.size(); i++) {
            if (s1[i] == s2[i]) continue;

            for (int j = 0; j < s1.size(); j++) {
                if (s1[j] == s2[j]) continue;
                if (s2[i] == s1[j]) {
                    char temp = s1[i];
                    s1[i] = s2[i];
                    s1[j] = temp;
                    res.insert(s1);
                    s1[i] = temp;
                    s1[j] = s2[i];
                }
            }
            break;
        }

        return res;
    }

    int kSimilarity(string s1, string s2) {
        queue<string> q;
        unordered_set<string> visited;
        int count = 0;

        q.push(s1);
        visited.insert(s1);

        while (q.size() > 0) {
            int sz = q.size();

            for (int i = 0; i < sz; i++) {
                string curr = q.front(); q.pop();

                if (curr.compare(s2) == 0) return count;

                unordered_set<string> nei = getNext(curr, s2);

                for (auto& x : nei) {
                    if (visited.count(x) == 0) {
                        visited.insert(x);
                        q.push(x);
                    }
                }
            }
            count++;
        }

        return -1;
    }
};
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