<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:
    def findRotateSteps(self, ring: str, key: str) -> int:
        
        distances_right = defaultdict(lambda: [len(ring)] * len(ring))
        distances_left = defaultdict(lambda: [len(ring)] * len(ring))

        def create_distances(ch: str, idx: int):
            for i in range(len(ring)):
                place_index = (idx - i) % len(ring)
                distances_right[ch][place_index] = min(distances_right[ch][place_index], i)
            for i in range(len(ring)):
                place_index = (idx + i) % len(ring)
                distances_left[ch][place_index] = min(distances_left[ch][place_index], i)
        
        current_idx = 0
        steps = 0

        for i, ch in enumerate(ring):
            create_distances(ch, i)

        print(distances_right["o"])
        print(distances_left["o"])

        @cache
        def dp(ring_idx: int, key_idx: str):
            if key_idx == len(key):
                return 0
            
            r_dist = distances_right[key[key_idx]][ring_idx]
            r_next_idx = (ring_idx + r_dist) % len(ring)
            clockwise_total = 1 + r_dist + dp(r_next_idx, key_idx + 1)

            l_dist = distances_left[key[key_idx]][ring_idx]
            l_next_idx = (ring_idx - l_dist) % len(ring)
            counterclockwise_total = 1 + l_dist + dp(l_next_idx, key_idx + 1)

            return min(clockwise_total, counterclockwise_total)
            

        return dp(0, 0)







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