<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 ringLen = ring.length();
        int keyLen = key.length();
        
        // HashMap to store the indices of occurrences of each character in the ring
        unordered_map<char, vector<int>> characterIndices;
        for (int i = 0; i < ring.length(); i++) {
            char ch = ring[i];
            characterIndices[ch].push_back(i);
        }

        // Initialize the heap (priority queue) with the starting point
        // Each element of the heap is a vector of integers representing:
        // totalSteps, ringIndex, keyIndex
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> heap;
        heap.push({0, 0, 0});

        // HashSet to keep track of visited states
        unordered_set<string> seen;
        
        // Spell the keyword using the metal dial
        int totalSteps = 0;
        while (!heap.empty()) {
            // Pop the element with the smallest total steps from the heap
            vector<int> state = heap.top();
            heap.pop();
            totalSteps = state[0];
            int ringIndex = state[1];
            int keyIndex = state[2];
            
            // We have spelled the keyword
            if (keyIndex == keyLen) {
                break;
            }
            
            // Continue if we have visited this character from this position in ring before
            string currentState = to_string(ringIndex) + "-" + to_string(keyIndex);
            if (seen.count(currentState)) {
                continue;
            }

            // Otherwise, add this pair to the visited list
            seen.insert(currentState);
            
            // Add the rest of the occurrences of this character in ring to the heap
            for (int nextIndex : characterIndices[key[keyIndex]]) {
                heap.push({totalSteps + countSteps(ringIndex, nextIndex, ringLen), 
                            nextIndex, keyIndex + 1});
            }
        }
        
        // Return the total steps and add keyLen to account for 
        // pressing the center button for each character in the keyword
        return totalSteps + keyLen;
    }

private:
    // Find the minimum steps between two indexes of ring
    int countSteps(int curr, int next, int ringLength) {
        int stepsBetween = abs(curr - next);
        int stepsAround = ringLength - stepsBetween;
        return min(stepsBetween, stepsAround);
    }
};
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