<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:
    unordered_map<int,int> dijsktras(int src, auto& graph) {
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq;
        pq.push({0, src});
        
        unordered_set<int> visited;
        unordered_map<int, int> min_cost_map;
        int minCost = INT_MAX;
        while (!pq.empty()) {
            int dist = pq.top().first;
            int u = pq.top().second;
            pq.pop();
            
            if (visited.find(u) == visited.end()) {
                // cout<<((char)u)<<" "<<dist<<endl;
                visited.insert(u);
                min_cost_map[u] = dist;
                for (auto& [v, v_dist]: graph[u]) {
                    int cost = dist+v_dist;
                    // cout<<(char)u<<"->"<<(char)v<<"=="<<(char)tar<<" "<<cost<<endl;
                    pq.push({cost, v});
                }
            }
        }
        
        return min_cost_map;
    }
    
    long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
        // Approach: graph single source shortest path (Dijsktra's)
        // n -> len(original), m -> len(source)
        // Time: O(n+m.n) ~ O(m.n) | n -> creating graph, m.n -> for each char in src(m), do dijsktras's(n-edges) [worst case]
        // Space: O(3n+n+n+n.n) ~ O(n^2) | 3n -> graph, n -> visited set, n -> pq, n.n -> min_cost_maps
        
        int m = source.size();
        int n = original.size();
        
        // create directed graph
        unordered_map<int, vector<pair<int,int>>> graph;
        for (int i=0; i<n; i++)
            graph[original[i]].push_back({changed[i], cost[i]});
        
        // print graph
        // for (auto& [u, v]: graph) {
        //     cout<<u<<"->[";
        //     for (auto& _v: v)
        //         cout<<"["<<_v.first<<","<<_v.second<<"],";
        //     cout<<"]"<<endl;
        // }
        
        // check each character in source and target string
        // if diff perform dijsktra's -> TLE
        long long minCost = 0;
        // int _cost = 0;
        // for (int i=0; i<m; i++) {
        //     if (source[i] != target[i]) {
        //         _cost = dijsktras(source[i], target[i], graph);
        //         // if cost is -1, can't convert src -> target
        //         if (_cost == -1) return -1;
        //         minCost += _cost;
        //     }
        // }
        
        // Optimization: Watched Neetcode youtube
        // since there are only 26 chars, so create 
        // hashmap that contains cost to each char from
        // each unique char in source
        unordered_map<int,unordered_map<int,int>> min_cost_maps;
        for (int i=0; i<m; i++) {
            if (source[i] != target[i]) {
                // the dijstras from source is not computed, then compute
                if (min_cost_maps.find(source[i]) == min_cost_maps.end()) {
                    min_cost_maps[source[i]] = dijsktras(source[i], graph);
                }
                // if already computed then use
                if (min_cost_maps[source[i]].find(target[i]) == min_cost_maps[source[i]].end())
                    return -1;
                minCost += min_cost_maps[source[i]][target[i]];
            }
        }
        
        // return minCost
        return minCost;
    }
};
159146