<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) {
        int dp[ring.size()][key.size()];
        map<char, vector<int>> table;
        vector<int> path;

        for(int i=0;i<ring.size();i++){
            table[ring[i]].emplace_back(i);
            if(ring[i] == key[key.size() - 1]){
                dp[i][key.size()-1] = 1;
            }
        }

        for(int index = key.size() - 2; index>-1; index--){
            vector<int> pos = table[key[index]];
            vector<int> next_pos = table[key[index+1]];
            for(int i=0;i<pos.size();i++){
                int move_step = INT_MAX, next = 0, cost = INT_MAX;
                for(int j=0;j<next_pos.size();j++){
                    int tmp = GetMinRotateNum(pos[i], next_pos[j], ring.size());
                    if(cost > (tmp+dp[next_pos[j]][index + 1])){
                        move_step = tmp;
                        next = next_pos[j];
                        cost = tmp+dp[next_pos[j]][index + 1];
                    }
                }
                dp[pos[i]][index] = 1 + move_step + dp[next][index + 1];
            }
        }

        int ans = -1;
        if(ring[0] != key[0]){
            vector<int> next_pos = table[key[0]]; 
            int move_step = INT_MAX, next = -1, cost = INT_MAX;
            for(int i=0;i<next_pos.size();i++){
                int tmp = GetMinRotateNum(0, next_pos[i], ring.size());
                if(cost > (tmp+dp[next_pos[i]][0])){
                    move_step = tmp;
                    next = next_pos[i];
                    cost = tmp+dp[next_pos[i]][0];
                }
            }
            ans = move_step + dp[next][0];
        }else{
            ans = dp[0][0];
        }     
        
        return ans;
    }

    int GetMinRotateNum(int src, int dst, int length){
        int a = (src - dst + length)%length;
        int b = (dst - src + length)%length;
        return a > b ? b : a;
    }
};
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