<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> findMyNeighbour(string temp1,string actual,unordered_map<string, int>&freq)
    {   
        
        vector<string>dump;
        int i = 0;
        for(i=0;i<temp1.size() && temp1[i]==actual[i] ; i++);

        for(int j=i;j<temp1.size();j++)
        {
            if(temp1[j]==actual[i])
            {
                char c = temp1[i];
                temp1[i] = temp1[j];
                temp1[j] = c;
                if(freq.find(temp1)==freq.end())
                {
                    freq[temp1]++;
                    dump.push_back(temp1);
                }

                temp1[j] = temp1[i];
                temp1[i] = c;
            }
        }

        return dump;
    }
    int kSimilarity(string s1, string s2) {
        int n = s1.size();
        if(s1==s2) return 0;
        unordered_map<string, int> freq;
        freq[s1]++;
        queue<pair<int, string>> q1;
        queue<pair<int, string>> q2;
        q1.push({0, s1});
        while (1) {

            if (!q1.empty()) {
                while (!q1.empty()) {
                    auto it = q1.front();
                    q1.pop();
                    string g1 = it.second;
                    int step = it.first;
                    vector<string> str_temp = findMyNeighbour(g1, s2, freq);
                    for (int i = 0; i < str_temp.size(); i++) {
                        if (str_temp[i] == s2) {
                            return step + 1;
                        }
                        q2.push({step + 1, str_temp[i]});
                    }
                }

            } else {
                while (!q2.empty()) {
                    auto it = q2.front();
                    q2.pop();
                    string g2 = it.second;
                    int step = it.first;
                    vector<string> str_temp = findMyNeighbour(g2, s2, freq);
                    for (int i = 0; i < str_temp.size(); i++) {
                        if (str_temp[i] == s2) {
                            return step + 1;
                        }
                        q1.push({step + 1, str_temp[i]});
                    }
                }
            }
        }

        return -1;
    }
};
118814