<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 {
private:
    map<string, map<string, int>> memo;
public:
    int kSimilarity(string s1, string s2) {
        if (memo[s1].contains(s2))
            return memo[s1][s2];
        // cout << s1 << ", " << s2 << endl;
        int n = s1.size();
        for (int i = 0; i < n; ++i) {
            if (s1[i] == s2[i])
                continue;
            int minIdx = INT_MAX;
            for (int j = i + 1; j < n; ++j) {
                if (s1[j] == s2[j])
                    continue;
                if (s1[j] != s2[i])
                    continue;
                swap(s1[i], s1[j]);
                minIdx = min(minIdx, 1 + kSimilarity(s1.substr(i + 1), s2.substr(i + 1)));
                swap(s1[i], s1[j]);
                if (s1[i] == s2[j] && s1[j] == s2[i]) break;
            }
            memo[s1][s2] = minIdx;
            return minIdx;
        }
        return 0;
    }
};
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