<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>

#define mod 1000000007
class Solution {
public:
    long long minimumCost(string source, string target, vector<char>& original, vector<char>& changed, vector<int>& cost) {
        //assume all characters as nodes; cost v given h ek node se dusre p jane ki; 
        //ab dekho source m kis kisko badalna h, un sbpar dijkstra laga do, thus finding the shortest path to
        //go from that node to target node; 

        int n = 26;
        vector<vector<int>> edges(n, vector<int>(n, INT_MAX));
        for (int i = 0; i < n; i++) edges[i][i] = 0;
        
        //populating edges vector
        for (int i = 0, m = cost.size(); i < m; i++) {
            edges[original[i] - 'a'][changed[i] - 'a'] = (edges[original[i] - 'a'][changed[i] - 'a'] > cost[i]) ? cost[i] : edges[original[i] - 'a'][changed[i] - 'a'];
        }

        vector<int> doshi, doshi2;
        for (int i = 0, m = source.size(); i < m; i++) {
            if (source[i] != target[i]) {
                doshi.push_back(source[i] - 'a');
                doshi2.push_back(target[i] - 'a');
            }
        }

        vector<vector<int>> dijkstra(n, vector<int>(n, INT_MAX));
        vector<bool> done(n, 0);
        for (int i = 0, m = doshi.size(); i < m; i++) {
            if (done[doshi[i]]) continue;
            vector<int> dist(n, INT_MAX);
            vector<bool> visited(n, 0);
            int cEle, cCost, vCount;

            cEle = doshi[i];
            cCost = 0;
            vCount = 1;
            visited[doshi[i]] = 1;
            dist[doshi[i]] = 0;

            while (vCount < n) {
                for (int j = 0; j < n; j++) {
                    if (edges[cEle][j] == INT_MAX) continue;
                    dist[j] = min (dist[j], edges[cEle][j] + cCost);
                }
                for (int j = 0, x = INT_MAX; j < n; j++) {
                    if (visited[j]) continue;
                    if (dist[j] < x) {
                        x = dist[j];
                        cEle = j;
                        cCost = x;
                    }
                }
                vCount++;
                visited[cEle] = 1;
            }

            for (int j = 0; j < n; j++) dijkstra[doshi[i]][j] = dist[j];
            done[doshi[i]] = 1;
        }

        long long ans = 0;
        for (int i = 0, m = doshi.size(); i < m; i++) {
            if (dijkstra[doshi[i]][doshi2[i]] == INT_MAX) return -1;
            ans += (dijkstra[doshi[i]][doshi2[i]]);
        }
        return ans;
    }
};
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