<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:
    // Memoization table
    unordered_map<string, int> memo;

    // making changes in string 2
    int dfs(string &s1, string &s2, int cur) {
        if (cur == s1.size()) return 0;
        if (memo.count(s2)) return memo[s2];

        int ans = INT_MAX;

        if (s1[cur] == s2[cur]) {
            return memo[s2] = dfs(s1, s2, cur + 1);
        }

        for (int i = cur + 1; i < s1.size(); i++) {
            if (s2[i] == s1[cur] && s2[i] != s1[i]) {
                swap(s2[i], s2[cur]);
                ans = min(ans, dfs(s1, s2, cur + 1));
                swap(s2[i], s2[cur]); // backtrack
            }
        }
        return memo[s2] = 1 + ans;
    }

    int kSimilarity(string s1, string s2) {
        string a = "", b = "";

        for (int i = 0; i < s1.length(); i++) {
            if (s1[i] != s2[i]) {
                a += s1[i];
                b += s2[i];
            }
        }

        return dfs(a, b, 0);
    }
};

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