<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 {
public:
    
long long getCost(vector<vector<pair<char, int>>>& graph, int start, int end) {
    // Initialize distances with infinity
    vector<long long> dist(26, LLONG_MAX);
    dist[start] = 0;

    // Queue for BFS
    priority_queue<int> q;
    q.push(start);

    while (!q.empty()) {
        int node = q.top();
        q.pop();

        // Process all adjacent nodes
        for (const auto& [neighbor, edgeCost] : graph[node]) {
            long long newCost = dist[node] + edgeCost;
            // If a cheaper path is found
            if (newCost < dist[neighbor]) {
                dist[neighbor] = newCost;
                q.push(neighbor);  // Enqueue the neighbor
            }
        }
    }

    // Return the minimum cost to reach `end` node, or -1 if unreachable
    return (dist[end] == LLONG_MAX) ? -1 : dist[end];
}

long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
    long long ans = 0;
    unordered_map<long long, long long> costMap;
    vector<vector<pair<char, int>>> g(26);
    for(int i = 0; i < cost.size(); ++i) {   //Create the weighted directed graph from the given data
        g[original[i]-'a'].push_back({changed[i]-'a', cost[i]});  
    }
    for(int i = 0; i < source.size(); ++i){  // now check for the cost to make both string equal
        if(source[i] == target[i]) continue;
        long long t = 0, hs = (source[i] - 'a')*100 + target[i] - 'a';
        if(costMap.find(hs) != costMap.end()) t = costMap[hs];   // first check in hash map if the pair is already present or not. If available then use it
        else {                               // if cost was not yet calculated, then get the cost and update the hash map as well.
            t = getCost(g, source[i] - 'a', target[i] - 'a');  
            costMap[hs] = t; 
        }  
        if(t == -1) return -1;
        ans += t;
    }
    return ans;
}};
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