<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:
    struct cmp {
        bool operator()(pair<int, int> &x, pair<int, int> &y) {
            return x.second > y.second;
        }  
    };
    
    unordered_map<int, int> getCost(int x, unordered_map<int,vector<pair<int, int>>> & data) {
        priority_queue<pair<int, int>, vector<pair<int, int>>, cmp> pq;
        
        unordered_map<int, int> cost;
        
        pq.push(make_pair(x, 0));
        
        while(!pq.empty()) {
            auto it = pq.top();
            pq.pop();
            
            if(cost.find(it.first) != cost.end()) {
                continue;
            }
            cost[it.first] = it.second;
            //cout<<data[it.first].size()<<endl;
            for(auto item: data[it.first]) {
                //cout<<"   = "<<item.first<<" "<<item.second<<endl;
                if(cost.find(item.first) != cost.end() && cost[item.first] > (it.second+item.second)) {
                    continue;                   
                }
                
                pq.push(make_pair(item.first, it.second + item.second));
            }    
        }
        return cost;
    }
    
    long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
        
        int n = original.size();

        unordered_map<int, vector<pair<int, int>>> graph;
        
        unordered_map<int, unordered_map<int, int>> result;
        for(int i = 0; i<n; i++) {
            graph[original[i] - 'a'].push_back(make_pair(changed[i] - 'a', cost[i]));
        }
        
        for(auto it: graph) {
            result[it.first] = getCost(it.first, graph);
            // for(auto j: result[it.first]) {
            //     cout<<"     "<<j.first<<" "<<j.second<<endl;
            // }
            //cout<<" "<<result[it.first].size()<<endl;
        }
        long long ans = 0;
        for(int i = 0; i<source.size(); i++) {
            if(source[i] == target[i]) continue;
            if(result[source[i] - 'a'].find(target[i]-'a') == result[source[i] - 'a'].end()) {
                return -1;
            }
            ans += result[source[i]-'a'][target[i]-'a'];
        }
        return ans;
        
    }
};
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