<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:
        len_ring = len(ring)

        # map chars -> indices in the ring
        idx_map = defaultdict(list)
        for i, ch in enumerate(ring):
            idx_map[ch].append(i)

        # pos 0, step 0
        prev_dp = {0: 0}
        #print(f"init: {prev_dp}\n")

        for idx, ch in enumerate(key):
            current_dp = {}
            cur_positions = idx_map[ch]
            #print(f"process char '{ch}' at key index {idx}")
            #print(f"positions of '{ch}' in ring: {cur_positions}")

            for cur_pos in cur_positions:
                min_steps = float('inf')
                for prev_pos, prev_steps in prev_dp.items():
                    # distance from prev -> current
                    delta = abs(cur_pos - prev_pos)
                    dist = min(delta, len_ring - delta)
                    steps = prev_steps + dist + 1

                    if steps < min_steps:
                        min_steps = steps

                current_dp[cur_pos] = min_steps
                #print(f"  min steps to reach position {cur_pos}: {min_steps}")

            prev_dp = current_dp
            #print(f"update dp: {prev_dp}\n")

        result = min(prev_dp.values())
        #print(f"min steps to spell the key: {result}")
        return result
16300