<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>

from typing import List

class Solution:
    def kSimilarity(self, s1: str, s2: str) -> int:
        memo = {}
        
        def dfs(s1, s2, i):
            if i == len(s1):
                return 0
            if (s1, s2, i) in memo:
                return memo[(s1, s2, i)]
            if s1[i] == s2[i]:
                ans = dfs(s1, s2, i + 1)
            else:
                indices = [j for j in range(i + 1, len(s1)) if s1[j] == s2[i]]
                ans = float('inf')
                for j in indices:
                    s1_list = list(s1)
                    s1_list[i], s1_list[j] = s1_list[j], s1_list[i]
                    s1_new = ''.join(s1_list)
                    ans = min(ans, dfs(s1_new, s2, i + 1) + 1)
            memo[(s1, s2, i)] = ans
            return ans
        
        return dfs(s1, s2, 0)
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