<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 characters. You are also given two <strong>0-indexed</strong> string arrays <code>original</code> and <code>changed</code>, and an integer array <code>cost</code>, where <code>cost[i]</code> represents the cost of converting the string <code>original[i]</code> to the string <code>changed[i]</code>.</p>

<p>You start with the string <code>source</code>. In one operation, you can pick a <strong>substring</strong> <code>x</code> from the string, and change it to <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>. You are allowed to do <strong>any</strong> number of operations, but any pair of operations must satisfy <strong>either</strong> of these two conditions:</p>

<ul>
	<li>The substrings picked in the operations are <code>source[a..b]</code> and <code>source[c..d]</code> with either <code>b &lt; c</code> <strong>or</strong> <code>d &lt; a</code>. In other words, the indices picked in both operations are <strong>disjoint</strong>.</li>
	<li>The substrings picked in the operations are <code>source[a..b]</code> and <code>source[c..d]</code> with <code>a == c</code> <strong>and</strong> <code>b == d</code>. In other words, the indices picked in both operations are <strong>identical</strong>.</li>
</ul>

<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</em>. <em>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 &quot;abcd&quot; to &quot;acbe&quot;, do the following operations:
- Change substring source[1..1] from &quot;b&quot; to &quot;c&quot; at a cost of 5.
- Change substring source[2..2] from &quot;c&quot; to &quot;e&quot; at a cost of 1.
- Change substring source[2..2] from &quot;e&quot; to &quot;b&quot; at a cost of 2.
- Change substring source[3..3] from &quot;d&quot; to &quot;e&quot; 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;abcdefgh&quot;, target = &quot;acdeeghh&quot;, original = [&quot;bcd&quot;,&quot;fgh&quot;,&quot;thh&quot;], changed = [&quot;cde&quot;,&quot;thh&quot;,&quot;ghh&quot;], cost = [1,3,5]
<strong>Output:</strong> 9
<strong>Explanation:</strong> To convert &quot;abcdefgh&quot; to &quot;acdeeghh&quot;, do the following operations:
- Change substring source[1..3] from &quot;bcd&quot; to &quot;cde&quot; at a cost of 1.
- Change substring source[5..7] from &quot;fgh&quot; to &quot;thh&quot; at a cost of 3. We can do this operation because indices [5,7] are disjoint with indices picked in the first operation.
- Change substring source[5..7] from &quot;thh&quot; to &quot;ghh&quot; at a cost of 5. We can do this operation because indices [5,7] are disjoint with indices picked in the first operation, and identical with indices picked in the second operation.
The total cost incurred is 1 + 3 + 5 = 9.
It can be shown that this is the minimum possible cost.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> source = &quot;abcdefgh&quot;, target = &quot;addddddd&quot;, original = [&quot;bcd&quot;,&quot;defgh&quot;], changed = [&quot;ddd&quot;,&quot;ddddd&quot;], cost = [100,1578]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It is impossible to convert &quot;abcdefgh&quot; to &quot;addddddd&quot;.
If you select substring source[1..3] as the first operation to change &quot;abcdefgh&quot; to &quot;adddefgh&quot;, you cannot select substring source[3..7] as the second operation because it has a common index, 3, with the first operation.
If you select substring source[3..7] as the first operation to change &quot;abcdefgh&quot; to &quot;abcddddd&quot;, you cannot select substring source[1..3] as the second operation because it has a common index, 3, with the first operation.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= source.length == target.length &lt;= 1000</code></li>
	<li><code>source</code>, <code>target</code> consist only of lowercase English characters.</li>
	<li><code>1 &lt;= cost.length == original.length == changed.length &lt;= 100</code></li>
	<li><code>1 &lt;= original[i].length == changed[i].length &lt;= source.length</code></li>
	<li><code>original[i]</code>, <code>changed[i]</code> consist only of lowercase English characters.</li>
	<li><code>original[i] != changed[i]</code></li>
	<li><code>1 &lt;= cost[i] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def minimumCost(self, s: str, t: str, original: List[str], changed: List[str], cost: List[int]) -> int:
      class Node:
        def __init__(self):
          self.is_word = False
          self.idx = -1 # no word here
          self.child = {}

      def add_word(root, s, idx):
        temp = root

        # Add word backward
        for i in range(len(s)-1, -1, -1):
          char = s[i]
          if char not in temp.child:
            temp.child[char] = Node()
          temp = temp.child[char]

        temp.idx = idx
      #---------------------------------------------------------
      INF = float('inf')
      dist = defaultdict(lambda: INF)
      v_to_idx = defaultdict(int)
      N, M = len(s), len(original)
      trie = Node()

      # Construct trie and graph
      for i in range(M):
        oi, ci = original[i], changed[i]

        # Map substr to idx and add (s, idx of s) to trie
        if oi not in v_to_idx:
          v_to_idx[oi] = len(v_to_idx)
          add_word(trie, oi, v_to_idx[oi])

        if ci not in v_to_idx:
          v_to_idx[ci] = v = len(v_to_idx)
          add_word(trie, ci, v_to_idx[ci])

        u, v = v_to_idx[oi], v_to_idx[ci]
        # Dup edges, take the min cost
        dist[(u, v)] = min(dist[(u, v)], cost[i])

      #--------------------------------------------------------
      V = len(v_to_idx)

      # Floyd-Warshall to find shortest path between all (u, v)
      for k in range(V):
        for i in range(V):
          # no edge (i, k)
          if dist[(i, k)] == INF:
            continue

          for j in range(V):
            # (k, j) must exists and intermediate path is shorter.
            if dist[(k, j)] != INF and dist[(i,k)]+dist[(k,j)] < dist[(i,j)]:
              dist[(i,j)] = dist[(i,k)] + dist[(k,j)]

      #--------------------------------------------------------
      result = 0
      dp = [INF for i in range(N)]

      for i in range(N):
        temp_s = trie
        temp_t = trie
        # traverse trie starting at parallel s[i] and t[i]
        for j in range(i, -1, -1):
          # either s[j -> i] no path or t[j -> i] no path
          if s[j] not in temp_s.child or t[j] not in temp_t.child:
            break

          temp_s, temp_t = temp_s.child[s[j]], temp_t.child[t[j]]

          # we found a s[j->i] and t[j->i]
          if temp_s.idx != -1 and temp_t.idx != -1:
            u, v = temp_s.idx, temp_t.idx
            dp[i] = min(dp[i], (dp[j-1] if j-1>=0 else 0) + dist[(u,v)])

        if s[i] == t[i]:
          dp[i] = min(dp[i-1] if i-1>=0 else 0, dp[i])

      return dp[-1] if dp[-1] < INF else -1
4009