<p>You are given two <strong>0-indexed</strong> strings <code>source</code> and <code>target</code>, both of length <code>n</code> and consisting of <strong>lowercase</strong> English letters. You are also given two <strong>0-indexed</strong> character arrays <code>original</code> and <code>changed</code>, and an integer array <code>cost</code>, where <code>cost[i]</code> represents the cost of changing the character <code>original[i]</code> to the character <code>changed[i]</code>.</p>

<p>You start with the string <code>source</code>. In one operation, you can pick a character <code>x</code> from the string and change it to the character <code>y</code> at a cost of <code>z</code> <strong>if</strong> there exists <strong>any</strong> index <code>j</code> such that <code>cost[j] == z</code>, <code>original[j] == x</code>, and <code>changed[j] == y</code>.</p>

<p>Return <em>the <strong>minimum</strong> cost to convert the string </em><code>source</code><em> to the string </em><code>target</code><em> using <strong>any</strong> number of operations. If it is impossible to convert</em> <code>source</code> <em>to</em> <code>target</code>, <em>return</em> <code>-1</code>.</p>

<p><strong>Note</strong> that there may exist indices <code>i</code>, <code>j</code> such that <code>original[j] == original[i]</code> and <code>changed[j] == changed[i]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> source = &quot;abcd&quot;, target = &quot;acbe&quot;, original = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;c&quot;,&quot;e&quot;,&quot;d&quot;], changed = [&quot;b&quot;,&quot;c&quot;,&quot;b&quot;,&quot;e&quot;,&quot;b&quot;,&quot;e&quot;], cost = [2,5,5,1,2,20]
<strong>Output:</strong> 28
<strong>Explanation:</strong> To convert the string &quot;abcd&quot; to string &quot;acbe&quot;:
- Change value at index 1 from &#39;b&#39; to &#39;c&#39; at a cost of 5.
- Change value at index 2 from &#39;c&#39; to &#39;e&#39; at a cost of 1.
- Change value at index 2 from &#39;e&#39; to &#39;b&#39; at a cost of 2.
- Change value at index 3 from &#39;d&#39; to &#39;e&#39; at a cost of 20.
The total cost incurred is 5 + 1 + 2 + 20 = 28.
It can be shown that this is the minimum possible cost.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> source = &quot;aaaa&quot;, target = &quot;bbbb&quot;, original = [&quot;a&quot;,&quot;c&quot;], changed = [&quot;c&quot;,&quot;b&quot;], cost = [1,2]
<strong>Output:</strong> 12
<strong>Explanation:</strong> To change the character &#39;a&#39; to &#39;b&#39; change the character &#39;a&#39; to &#39;c&#39; at a cost of 1, followed by changing the character &#39;c&#39; to &#39;b&#39; at a cost of 2, for a total cost of 1 + 2 = 3. To change all occurrences of &#39;a&#39; to &#39;b&#39;, a total cost of 3 * 4 = 12 is incurred.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> source = &quot;abcd&quot;, target = &quot;abce&quot;, original = [&quot;a&quot;], changed = [&quot;e&quot;], cost = [10000]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It is impossible to convert source to target because the value at index 3 cannot be changed from &#39;d&#39; to &#39;e&#39;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= source.length == target.length &lt;= 10<sup>5</sup></code></li>
	<li><code>source</code>, <code>target</code> consist of lowercase English letters.</li>
	<li><code>1 &lt;= cost.length == original.length == changed.length &lt;= 2000</code></li>
	<li><code>original[i]</code>, <code>changed[i]</code> are lowercase English letters.</li>
	<li><code>1 &lt;= cost[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>original[i] != changed[i]</code></li>
</ul>

const static auto speedup = [] () {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    
    return 0;
} ();

class Solution {
public:
    long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
        unordered_map<char, unordered_map<char, int>> g;
        for (int i = 0; i < original.size(); ++i) {
           char src = original[i];
           char dst = changed[i];
           int dist = cost[i];

           if (src == dst) continue;
           if (g[src][dst] != 0 && g[src][dst] < dist) continue;
           g[src][dst] = dist; 
        }

        long long res = 0;

        unordered_map<char, unordered_map<char, long long>> memo;
        for (auto c: unordered_set(begin(source), end(source))) {
            memo[c] = helper(g, c);
        }

        for (int i = 0; i < source.size(); ++i) {
            char src = source[i], dst = target[i];
            if (memo[src].count(dst) == 0) return -1;
            res += memo[src][dst];
        }

        return res;
    }

private:
    unordered_map<char, long long> helper(unordered_map<char, unordered_map<char, int>>& g, int src) {
        unordered_map<char, long long> res;
        
        std::priority_queue<pair<long long, char>, vector<pair<long long, char>>, greater<pair<long long, char>>> pq;
        unordered_set<char> visited;
        pq.push(make_pair(0L, src));

        while (!pq.empty()) {
            auto top = pq.top(); pq.pop();
            if (visited.count(top.second)) {
                continue;
            }
            visited.insert(top.second);
            res[top.second] = top.first;

            for (auto [next, dist]: g[top.second]) {
                pq.push(make_pair(dist + top.first, next));
            }
        }

        return res;
    }
    // long long helper(unordered_map<char, unordered_map<char, int>>& g, int src, int dst) {
    //     std::priority_queue<pair<long long, char>, vector<pair<long long, char>>, greater<pair<long long, char>>> pq;
    //     unordered_set<char> visited;
    //     pq.push(make_pair(0L, src));

    //     while (!pq.empty()) {
    //         auto top = pq.top(); pq.pop();
    //         if (visited.count(top.second)) {
    //             continue;
    //         }
    //         visited.insert(top.second);
    //         if (top.second == dst) return top.first;

    //         for (auto [next, dist]: g[top.second]) {
    //             pq.push(make_pair(dist + top.first, next));
    //         }
    //     }

    //     return -1;
    // }
};
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