<p>In the video game Fallout 4, the quest <strong>&quot;Road to Freedom&quot;</strong> requires players to reach a metal dial called the <strong>&quot;Freedom Trail Ring&quot;</strong> and use the dial to spell a specific keyword to open the door.</p>

<p>Given a string <code>ring</code> that represents the code engraved on the outer ring and another string <code>key</code> that represents the keyword that needs to be spelled, return <em>the minimum number of steps to spell all the characters in the keyword</em>.</p>

<p>Initially, the first character of the ring is aligned at the <code>&quot;12:00&quot;</code> direction. You should spell all the characters in <code>key</code> one by one by rotating <code>ring</code> clockwise or anticlockwise to make each character of the string key aligned at the <code>&quot;12:00&quot;</code> direction and then by pressing the center button.</p>

<p>At the stage of rotating the ring to spell the key character <code>key[i]</code>:</p>

<ol>
	<li>You can rotate the ring clockwise or anticlockwise by one place, which counts as <strong>one step</strong>. The final purpose of the rotation is to align one of <code>ring</code>&#39;s characters at the <code>&quot;12:00&quot;</code> direction, where this character must equal <code>key[i]</code>.</li>
	<li>If the character <code>key[i]</code> has been aligned at the <code>&quot;12:00&quot;</code> direction, press the center button to spell, which also counts as <strong>one step</strong>. After the pressing, you could begin to spell the next character in the key (next stage). Otherwise, you have finished all the spelling.</li>
</ol>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2018/10/22/ring.jpg" style="width: 450px; height: 450px;" />
<pre>
<strong>Input:</strong> ring = &quot;godding&quot;, key = &quot;gd&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong>
For the first key character &#39;g&#39;, since it is already in place, we just need 1 step to spell this character. 
For the second key character &#39;d&#39;, we need to rotate the ring &quot;godding&quot; anticlockwise by two steps to make it become &quot;ddinggo&quot;.
Also, we need 1 more step for spelling.
So the final output is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> ring = &quot;godding&quot;, key = &quot;godding&quot;
<strong>Output:</strong> 13
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= ring.length, key.length &lt;= 100</code></li>
	<li><code>ring</code> and <code>key</code> consist of only lower case English letters.</li>
	<li>It is guaranteed that <code>key</code> could always be spelled by rotating <code>ring</code>.</li>
</ul>

class Solution {
 public:
  int findRotateSteps(string ring, string key) {
    return dfs(ring, key, 0, {}) + key.length();
  }

 private:
  // Returns the number of rotates of ring to match key[index..n).
  int dfs(const string& ring, const string& key, int index,
          unordered_map<string, int>&& mem) {
    if (index == key.length())
      return 0;
    // Add the index to prevent duplication.
    const string hashKey = ring + to_string(index);
    if (const auto it = mem.find(hashKey); it != mem.cend())
      return it->second;

    int ans = INT_MAX;

    // For each ring[i] == key[index], we rotate the ring to match the ring[i]
    // with the key[index], then recursively match the newRing with the
    // key[index + 1..n).
    for (size_t i = 0; i < ring.length(); ++i)
      if (ring[i] == key[index]) {
        const int minRotates = min(i, ring.length() - i);
        const string& newRing = ring.substr(i) + ring.substr(0, i);
        const int remainingRotates =
            dfs(newRing, key, index + 1, std::move(mem));
        ans = min(ans, minRotates + remainingRotates);
      }

    return mem[hashKey] = ans;
  }
};
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