<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 {
    int n, m;
    map<string,int> map;
    int mod = 1000000007;
public:
    int kSimilarity(string s1, string s2) {
        string a, b;
        for(int i=0; i<s1.size(); i++){
            if(s1[i] != s2[i]){                 // remove all same char
                a = a + s1[i];
                b = b + s2[i];
            }
        }
        n = a.size();
        return backtrack(0, a, b);
    }
    int backtrack(int i, string s1, string s2) {
        if(i ==  n) {
            return 0;
        };
        if(s1[i] == s2[i]) return backtrack(i+1,s1, s2);   // skip if same char 
        if(map.find(s1)!=map.end()) return map[s1];
        int ans = INT_MAX;
        for(int idx =  i + 1; idx < n; ++idx){
            if(s1[idx]==s2[i] && s1[idx]!=s2[idx]){ 
                swap(s1[i], s1[idx]);
                int step = (1 + backtrack(i + 1, s1 , s2)) % mod;
                ans = min(step, ans);
                swap(s1[i], s1[idx]); // backtrack
            }         
        }
        map.insert({s1,ans});
    
        return ans;

    }
};
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