<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:
        forward_next = defaultdict(lambda: {})
        # forward_next = {}
        for o, c, cost_ in zip(original, changed, cost):
            # it is weird but the data does have multiple costs for a (o, c) pair
            if c not in forward_next[o] or cost_ < forward_next[o][c]:
                forward_next[o][c] = cost_
            # if o in forward_next:
            #     forward_next[o][0].append(c)
            #     forward_next[o][1].append(cost_)
            # else:
            #     forward_next[o] = ([c], [cost_])
        
        path_cache = {}
        path_last_search_cache = {}
        path_exhausted = set()
        def find_char_path(src_char, dst_char):
            nonlocal path_cache, path_last_search_cache
            if src_char in path_exhausted:
                if dst_char in path_cache[src_char]:
                    return path_cache[src_char][dst_char]
                else:
                    return None
            if src_char in path_cache and dst_char in path_cache[src_char]:
                return path_cache[src_char][dst_char]
            if src_char in path_last_search_cache:
                src_char_dict = path_cache[src_char]
                search_queue, visited = path_last_search_cache[src_char]
            else:
                src_char_dict = {}
                search_queue = []
                visited = set()
                heapq.heappush(search_queue, (0, src_char))
            while search_queue:
                cost, char = heapq.heappop(search_queue)
                if char in visited:
                    continue
                src_char_dict[char] = cost
                if char == dst_char:
                    heapq.heappush(search_queue, (cost, char))
                    path_last_search_cache[src_char] = (search_queue, visited)
                    path_cache[src_char] = src_char_dict
                    return cost
                visited.add(char)
                if char not in forward_next:
                    continue
                # next_chars, costs = forward_next[char]
                for next_char, cost_ in forward_next[char].items():
                # for next_char, cost_ in zip(next_chars, costs):
                    if next_char not in visited:
                        heapq.heappush(search_queue, (cost + cost_, next_char))
            path_exhausted.add(src_char)
            path_cache[src_char] = src_char_dict
            return None
        
        # path_cache = {}
        # def find_char_path(src_char):
        #     if src_char in path_cache:
        #         return
        #     ans_dict = {}
        #     search_queue = []
        #     visited = defaultdict(lambda: float('inf'))
        #     heapq.heappush(search_queue, (0, src_char))
        #     while search_queue:
        #         cost, char = heapq.heappop(search_queue)
        #         if char in ans_dict:
        #             continue
        #         ans_dict[char] = cost
        #         # if char not in forward_next:
        #         #     continue
        #         for next_char, cost_ in forward_next[char].items():
        #         # for next_char, cost_ in zip(forward_next[char][0], forward_next[char][1]):
        #             new_cost = cost + cost_
        #             if new_cost < visited[next_char]:
        #                 visited[next_char] = new_cost
        #                 heapq.heappush(search_queue, (cost + cost_, next_char))
        #     path_cache[src_char] = ans_dict
        
        ans = 0
        for src_char, dst_char in zip(source, target):
            if src_char == dst_char:
                continue
            # if src_char not in path_cache:
            #     find_char_path(src_char)
            # if dst_char in path_cache[src_char]:
            #     ans += path_cache[src_char][dst_char]
            # else:
            #     return -1
            
            if src_char in path_cache:
                if dst_char in path_cache[src_char]:
                    ans += path_cache[src_char][dst_char]
                    continue
                elif src_char in path_exhausted:
                    return -1
            
            src_char_cost = find_char_path(src_char, dst_char)
            if src_char_cost is None:
                return -1
            ans += src_char_cost
        
        return ans
715