<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:
    vector< string > getNeighbors(string s1, string s2) {
        
        vector< string > neighbors;

        int i = 0, n = s1.size();

        for (i = 0; i < n && s1[i] == s2[i]; i++) {}

        for (int j = i + 1; j < n; j++) {

            if (s1[j] == s2[i] && s1[j] != s2[j]) {

                swap(s1[j], s1[i]);
                neighbors.push_back(s1);
                swap(s1[j], s1[i]);
            }
        }

        return neighbors;
    }

    int kSimilarity(string s1, string s2) {

        queue<string> q;       
        q.push(s1);

        unordered_set< string > visited;
        visited.insert(s1);

        int steps = 0;

        while (true) {

            int size = q.size();
            for (int i = 0; i < size; i++) {

                string s = q.front();
                q.pop();

                if (s == s2)
                    return steps;

                vector< string > neighbors = getNeighbors(s, s2);
                for (string neighbor : neighbors) {

                    if (visited.find(neighbor) == visited.end()) {

                        q.push(neighbor);
                        visited.insert(neighbor);
                    }
                }
            }

            steps += 1;
        }

        return steps;
    }
};
59905