<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>

// class Solution {
//     using Graph = unordered_map<char, vector<pair<int, char>>>;
// public:
//     long long minimumCost(string source, string target, vector<char> &original,
//                           vector<char> &changed, vector<int> &cost) {
//         Graph graph;
//         for (int i = 0; i < original.size(); ++i) {
//             auto u = original[i];
//             auto v = changed[i];
//             int weight = cost[i];
//             graph[u].emplace_back(weight, v);
//         }
// unordered_map<char, unordered_map<char, int>> mmps;
//         auto djs = [&](char src) {
//             priority_queue<pair<int, char>, vector<pair<int, char>>, greater<>> pq;
//             pq.emplace(0, src);
//             unordered_map<char, int> mmp;
//             while (!pq.empty()) {
//                 char node = pq.top().second;
//                 int cost = pq.top().first;
//                 pq.pop();
//                 if (mmp.count(node)) continue;
//                 mmp[node] = cost;
//                 for (auto edges: graph[node]) {
//                     pq.emplace(cost + edges.first, edges.second);
//                 }
//             }
//             mmps[src]=mmp;
//         };

//         unordered_set<char> sets(source.begin(), source.end());
        
//         for (auto &e: sets) {
//              djs(e);
//         }

//         long long ans = 0;
//         for (int i = 0; i < source.size(); ++i) {
//             if (!mmps[source[i]].count(target[i])) {
//                 return -1;
//             }
//             ans += mmps[source[i]][target[i]];
//         }
//         return ans;
//     }
// };


class Solution {
    using Graph = unordered_map<char, vector<pair<int, char>>>;
public:
    long long minimumCost(string source, string target, vector<char> &original,
                          vector<char> &changed, vector<int> &cost) {
        Graph graph;
        unordered_map<char, unordered_map<char, int>> shortestPaths;

        // Build the graph
        for (int i = 0; i < original.size(); ++i) {
            char u = original[i];
            char v = changed[i];
            int weight = cost[i];
            graph[u].emplace_back(weight, v);
        }

        auto dijkstra = [&](char src) {
            if (shortestPaths.count(src)) return;
            
            priority_queue<pair<int, char>, vector<pair<int, char>>, greater<>> pq;
            pq.emplace(0, src);
            unordered_map<char, int> minCost;

            while (!pq.empty()) {
                auto [currCost, node] = pq.top();
                pq.pop();
                if (minCost.count(node)) continue;
                minCost[node] = currCost;
                for (auto& [weight, neighbor] : graph[node]) {
                    if (!minCost.count(neighbor)) {
                        pq.emplace(currCost + weight, neighbor);
                    }
                }
            }
            shortestPaths[src] = move(minCost);  // Move the map to avoid unnecessary copying
        };

        // Calculate the total minimum cost
        long long totalCost = 0;
        for (int i = 0; i < source.size(); ++i) {
            char src = source[i];
            char tgt = target[i];
            if (!shortestPaths.count(src)) {
                dijkstra(src);
            }
            if (!shortestPaths[src].count(tgt)) {
                return -1;
            }
            totalCost += shortestPaths[src][tgt];
        }

        return totalCost;
    }
};

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