<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 minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
        unordered_map<char, vector<char>>edges;
        //unordered_map<pair<char,char>, int>lowest_cost;
        //unordered_map<pair<char, char>, bool>lowest_cost_exists;

        int lowest_cost [26][26];
        bool lowest_cost_exists [26][26];
        int cur_cost[26][26];
        for(int i = 0; i<26; i++) {
            for(int j = 0; j<26; j++) {
                lowest_cost_exists[i][j] = false;
                lowest_cost[i][j] = 0;
                cur_cost[i][j] = INT_MAX;
            }
        }
        //unordered_map<pair<char,char>, int>cur_cost;
        for(int i = 0; i<original.size(); i++)  {
            cur_cost[original[i]-'a'][changed[i]-'a'] = min(cost[i],cur_cost[original[i]-'a'][changed[i]-'a']);
            edges[original[i]].push_back(changed[i]);
        }
        bool canDO = true;
        long long int ans = 0;

        for(int i = 0; i<source.size(); i++) {
            char s = source[i];
            char t = target[i];
            if(s==t) {
                continue;
            }
            //cout << "S: " << s << endl;
            pair<char, char> pst;
            pst.first = s;
            pst.second = t;
            if(lowest_cost_exists[s-'a'][t-'a']) {
                //cout << "lowest_cost_exists" << endl;
                ans += lowest_cost[s-'a'][t-'a'];
            }
            else  {
                priority_queue<pair<int,char>,vector<pair<int, char>>, greater<pair<int, char>>>pq;
                pq.push(make_pair(0,s));
                int arr[26];
                bool visisted[26];
                for(int i = 0; i<26; i++) {
                    arr[i] = INT_MAX;
                    visisted[i] = false;
                }
                visisted[s-'a'] =true;
                arr[s-'a'] = 0;

                while(!pq.empty()) {
                    pair<int,char> c = pq.top();
                    //cout << "PQ TOP: " << c.second << endl;
                    if(lowest_cost_exists[s-'a'][c.second-'a']) {
                        //cout << "lowest_cost_exists" << endl;
                        pq.pop();
                        continue;
                    }
                    visisted[c.second-'a'] = true;
                    lowest_cost[s-'a'][c.second-'a'] = c.first;
                    lowest_cost_exists[s-'a'][c.second-'a'] = true;
                    
                    pq.pop();
                    for(int  i= 0; i<edges[c.second].size(); i++) {
                        if(c.second=='c') {
                            //cout << "in for loop: neighbor: " << edges[c.second][i] <<  endl;
                        }
                        char neighbor = edges[c.second][i];
                        if(visisted[neighbor-'a']) {
                            //cout << "visited: " << neighbor << endl;
                            continue;
                        }
                        
                        if(arr[c.second-'a'] + cur_cost[c.second-'a'][neighbor-'a'] < arr[neighbor-'a']) {
                            arr[neighbor-'a'] = arr[c.second-'a'] + cur_cost[c.second-'a'][neighbor-'a'];
                            //cout << "updating arr" << endl;
                            //cout << c.second << " " << neighbor << " " << cur_cost[c.second-'a'][neighbor-'a'] << endl;
                        }
                        pq.push(make_pair(arr[neighbor-'a'], neighbor));
                    }
                }
                if(!lowest_cost_exists[s-'a'][t-'a']) {
                    canDO = false;
                    break;
                }
                else {
                    ////cout << s << " " << t << endl;
                    //cout << "lowest_Cost: " << lowest_cost[s-'a'][t-'a'] << endl;
                    ans += lowest_cost[s-'a'][t-'a'];
                }
            }
        }


        if(canDO) {return ans;} 
        {
            return -1;
        }

    }
};
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