<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>

# iterate over both source and target arrays and compare the values at same index. If they are different, need to run a dfs on every possibility of how you can convert a letter using the original/changed/cost arrays
# Seems like you can use dynamic programming to memoize the conversions. If you store {(curLetter, convertedLetter): cost} in memo array, then when you reach a combination of having a given letter and needing to convert to the target letter, you can just search memo array and return that, you know that it can be converted to the convertedLetter
# optimization problem, so that suggests dynamic programming or greedy usually.
# 
"""
class Solution:
    def __init__(self):
        self.memo = {} # {(curLetter, convertedLetter): cost}
        self.source = ""
        self.target = ""
        self.original = []
        self.changed = []
        self.cost = []

    def minimumCost(self, source: str, target: str, original: List[str], changed: List[str], cost: List[int]) -> int:
        self.source = source
        self.target = target
        self.original = original
        self.changed = changed
        self.cost = cost

        res = 0
        for i in range(len(self.source)):
            res += self.dfs(i, i)
        
        return res

    def dfs(self, srcIdx, tgtIdx):
        if srcIndex >= len(self.source) or tgtIdx >= len(self.target):
            return
        
        if (self.source[srcIdx], self.target[tgtIdx]) in self.memo:
            return self.memo[self.source[srcIdx], self.target[tgtIdx]]

        
        

        return min()
"""   

# approach 2
    # Dijkstras approach. Here you are looking at the minimum cost to reach a destination, and since we are doing something similar to a dfs to search through the possibilities, we want to follow the shortest path to reach the target cost. So usually you would have a list of [fromNode, toNode, weight], here you have a source letter, a target letter, and the weights are stored in the cost array. 
    # not bidirectional so you don't to append the inverse way
    # minHeap, need it structured to be (cost, curLetter) where curLetter will progress down to the target letter, and cost is the current cost of the path
        # want different minheap for each recursive call since you are calculating each path separately.
    # within dijkstras you want to use the minimum cost and return that 

class Solution:
    def __init__(self):
        self.memo = {} # {(curLetter, convertedLetter): cost}

    def minimumCost(self, source: str, target: str, original: List[str], changed: List[str], cost: List[int]) -> int:
        res = 0
        self.adjList = {}

        # populate adjacency list
        for i in range(len(original)):
            if original[i] not in self.adjList:
                self.adjList[original[i]] = []
            self.adjList[original[i]].append((changed[i], cost[i]))

        # iterate over each character in source
        for srcLetter, tgtLetter in zip(source, target): # zip gives you value not index
            costToConvert = self.dijkstra(srcLetter, tgtLetter)
            if costToConvert == -1:
                return -1
            res += costToConvert
        
        return res
    
    def dijkstra(self, startLetter, targetLetter):
        if (startLetter, targetLetter) in self.memo:
                return self.memo[(startLetter, targetLetter)]

        minHeap = [(0, startLetter)] # (curCost, curLetter)
        minCostMap = {startLetter: 0}

        while minHeap:
            curCost, curLetter = heappop(minHeap)
            
            if curLetter == targetLetter:
                self.memo[(startLetter, targetLetter)] = curCost
                return curCost

            if curCost > minCostMap[curLetter]: # not the lowest cost, don't use this
                continue

            if curLetter in self.adjList: # can be letters you visit that weren't a part of the original source char list
                for adjLetter, adjCost in self.adjList[curLetter]:
                    newCost = curCost + adjCost
                    if adjLetter not in minCostMap or (adjLetter in minCostMap and newCost < minCostMap[adjLetter]):
                        minCostMap[adjLetter] = newCost
                        heappush(minHeap, (newCost, adjLetter))

        return -1
    
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