<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) {
        
        if (s1 == s2) {
            return 0;
        }
        
        int res = 0;
        queue<string>q;
        unordered_set<string> visited;
        
        q.push(s1);
        visited.insert(s1);
        
        while(!q.empty()) {
            int n = q.size();
            while(n--) {
                
                string front = q.front();
                q.pop();
                
                if (front == s2) {
                    return res;
                }
                
                int i = 0;
                while(i < front.size() && front[i] == s2[i]) {
                    i++;
                }
                
                for (int j = i+1;j < front.size();j++) {
                    
                    if (front[j] != s2[j] && front[j] == s2[i]) {
                        swap(front[i],front[j]);
                        if (!visited.count(front)) {
                            q.push(front);
                            visited.insert(front);
                        }
                        swap(front[i],front[j]);
                    }
                }
            }
            ++res;
        }
        return res;
    }
};
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