<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:
    def minimumCost(self, source: str, target: str, original: List[str], changed: List[str], cost: List[int]) -> int:
    
        # # DEJKSTRA
        # def getCost(start, end, dist, graph, counted):
        #     if not counted[start]:
        #         dist_row = dist[start]
        #         visited = [False] * 26
        #         heap = [(0, start)]
        #         while heap:
        #             curCost, cur = heappop(heap)
        #             if visited[cur]:
        #                 continue
        #             visited[cur] = True
        #             for adj, adjCost in graph[cur]:
        #                 if curCost + adjCost < dist_row[adj]:
        #                     dist_row[adj] = curCost + adjCost
        #                     heappush(heap, (dist_row[adj], adj))
        #         counted[start] = True
        #     return dist[start][end]

        # numMap = {ch: ord(ch) - ord('a') for ch in string.ascii_lowercase}
        # dist = [[inf] * 26 for i in range(26)]
        # for i in range(26):
        #     dist[i][i] = 0
        # counted = [False] * 26
        # graph = defaultdict(list)
        # for i in range(len(cost)):
        #     graph[numMap[original[i]]].append((ord(changed[i]) - ord('a'), cost[i]))
        # ans = 0
        # for i in range(len(source)):
        #     if source[i] == target[i]:
        #         continue
        #     price = getCost(numMap[source[i]], numMap[target[i]], dist, graph, counted)
        #     if price == inf:
        #         return -1
        #     ans += price
        # return ans


        # # Floyd-Warshal
        # numMap = {ch: ord(ch) - ord('a') for ch in string.ascii_lowercase}
        # dist = [[inf] * 26 for i in range(26)]
        # for i in range(26):
        #     dist[i][i] = 0
        # for i in range(len(cost)):
        #     idx1, idx2 = numMap[original[i]], numMap[changed[i]]
        #     dist[idx1][idx2] = min(cost[i], dist[idx1][idx2])
        # for k in range(26):
        #     for i in range(26):
        #         for j in range(26):
        #             if dist[i][j] > dist[i][k] + dist[k][j]:
        #                 dist[i][j] = dist[i][k] + dist[k][j]
        # ans = 0
        # for i in range(len(source)):
        #     price = dist[numMap[source[i]]][numMap[target[i]]]
        #     if price == inf:
        #         return -1
        #     ans += price
        # return ans


        # # Bellman–Ford Algorithm
        # def getCost(start, end, dist, graph, counted):
        #     if not counted[start]:
        #         dist_row = dist[start]
        #         for _ in range(26):
        #             isIdle = True
        #             for cur in range(26):
        #                 for adj, adjCost in graph[cur]:
        #                     if dist_row[cur] + adjCost < dist_row[adj]:
        #                         dist_row[adj] = dist_row[cur] + adjCost
        #                         isIdle = False
        #             if isIdle:
        #                 break
        #         counted[start] = True
        #     return dist[start][end]

        # numMap = {ch: ord(ch) - ord('a') for ch in string.ascii_lowercase}
        # dist = [[inf] * 26 for i in range(26)]
        # for i in range(26):
        #     dist[i][i] = 0
        # counted = [False] * 26
        # graph = defaultdict(list)
        # for i in range(len(cost)):
        #     graph[numMap[original[i]]].append((ord(changed[i]) - ord('a'), cost[i]))
        # ans = 0
        # for i in range(len(source)):
        #     if source[i] == target[i]:
        #         continue
        #     price = getCost(numMap[source[i]], numMap[target[i]], dist, graph, counted)
        #     if price == inf:
        #         return -1
        #     ans += price
        # return ans


        # Shortest Path First Algorithm (SPFA) # improvement of Bellman–Ford
        def getCost(start, end, dist, graph, counted):
            if not counted[start]:
                dist_row = dist[start]
                qu = deque([start])
                while qu:
                    cur = qu.popleft()
                    for adj, adjCost in graph[cur]:
                        if dist_row[cur] + adjCost < dist_row[adj]:
                            dist_row[adj] = dist_row[cur] + adjCost
                            qu.append(adj)
                counted[start] = True
            return dist[start][end]

        numMap = {ch: ord(ch) - ord('a') for ch in string.ascii_lowercase}
        dist = [[inf] * 26 for i in range(26)]
        for i in range(26):
            dist[i][i] = 0
        counted = [False] * 26
        graph = defaultdict(list)
        for i in range(len(cost)):
            graph[numMap[original[i]]].append((numMap[changed[i]], cost[i]))
        ans = 0
        for i in range(len(source)):
            if source[i] != target[i]:
                price = getCost(numMap[source[i]], numMap[target[i]], dist, graph, counted)
                if price == inf:
                    return -1
                ans += price
        return ans
972