<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 n = ring.size();
        int m = key.size();
        vector<vector<int>> visited(n, vector<int>(m + 1, INT_MAX)); // Track minimum cost to reach (position, key_index)
        unordered_map<char, vector<int>> char_to_index; // Map each character to all its positions in ring

        // Preprocessing: map characters to their positions in the ring
        for (int i = 0; i < n; ++i) {
            char_to_index[ring[i]].push_back(i);
        }

        // BFS queue: (current_position, key_index, cost)
        queue<tuple<int, int, int>> q;
        q.push({0, 0, 0});
        visited[0][0] = 0;

        int min_steps = INT_MAX;

        while (!q.empty()) {
            auto [pos, key_idx, cost] = q.front();
            q.pop();

            // If we've spelled the entire key, update the minimum steps
            if (key_idx == m) {
                min_steps = min(min_steps, cost);
                continue;
            }

            char target_char = key[key_idx];
            
            // Explore all positions of the current target character
            for (int next_pos : char_to_index[target_char]) {
                // Calculate the rotation cost
                int steps = min(abs(next_pos - pos), n - abs(next_pos - pos)) + 1; // +1 for pressing the button
                int new_cost = cost + steps;

                // Only proceed if this is a cheaper way to reach this state
                if (new_cost < visited[next_pos][key_idx + 1]) {
                    visited[next_pos][key_idx + 1] = new_cost;
                    q.push({next_pos, key_idx + 1, new_cost});
                }
            }
        }

        return min_steps;
    }
};

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