<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:
    static int findRotateSteps(const string &ring, const string &key) {
        constexpr uint8_t ASIZE = 26, DSIZE = 100;
        const uint8_t r = ring.size(),
                      k = key.size();
        uint8_t pos[ASIZE][r+1u]; // todo shorten it
        for (uint8_t j = 0; j < ASIZE; j++)
            pos[j][0] = 0;
        for (uint8_t i = 0; i < r; i++) {
            uint8_t *const p = pos[ring[i]-'a'];
            p[++*p] = i;
        }
        uint16_t moves[2][DSIZE];
        fill(moves[0], moves[0] + DSIZE, USHRT_MAX);
        {
            const uint8_t *const p0 = pos[key[0]-'a'];
            for (uint8_t q = 1; q <= *p0; q++) {
                const uint8_t j = p0[q];
                moves[0][j] = min<uint8_t>(j, r - j) + 1u;
            }
        }
        for (uint8_t i = 1; i < k; i++) {
            const uint8_t *const p = pos[key[i-1u]-'a'],
                          *const c = pos[key[i]-'a'];
            fill(moves[i&1u], moves[i&1u] + DSIZE, USHRT_MAX);
            for (uint8_t q = 1; q <= *p; q++) {
                const uint8_t j = p[q];
                for (uint8_t d = 1; d <= *c; d++) {
                    const uint8_t e = c[d];
                    const uint8_t s = e > j ? e - j : j - e;
                    auto &m = moves[i&1u][e];
                    m = min<uint16_t>(m, moves[(i&1u)^1u][j] + min<uint8_t>(s, r - s) + 1u);
                }
            }
        }
        return *min_element(moves[(k&1u)^1u], moves[(k&1u)^1u] + r);
    }
};

auto init = []() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 'c';
}();
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