<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) {
        unordered_set<string> s;
        queue<tuple<string, int, int>> q;
        q.push({s1, 0, 0});
        s.insert(s1);
        while(q.size()){
            string t = get<0>(q.front());
            //cout<<"t: "<<t<<endl;
            int idx = get<1>(q.front());
            int step = get<2>(q.front());
            if(t == s2) return step;
            while(idx < t.size() && t[idx] == s2[idx]){
                idx ++;
            }
            int i;
            for(i = idx + 1 ; i < t.size() ; i ++){
                if(t[i] != s2[idx]) continue;
                //cout<<"i: "<<i<<endl;
                swap(t[idx], t[i]);
                //cout<<"t: "<<t<<endl;
                if(s.find(t) == s.end()){
                    q.push({t, idx + 1, step + 1});
                    s.insert(t);
                }
                swap(t[idx], t[i]);
            }
            q.pop();
        }
        return -1;
    }
};
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