<p>You are given a string <code>target</code>, an array of strings <code>words</code>, and an integer array <code>costs</code>, both arrays of the same length.</p>

<p>Imagine an empty string <code>s</code>.</p>

<p>You can perform the following operation any number of times (including <strong>zero</strong>):</p>

<ul>
	<li>Choose an index <code>i</code> in the range <code>[0, words.length - 1]</code>.</li>
	<li>Append <code>words[i]</code> to <code>s</code>.</li>
	<li>The cost of operation is <code>costs[i]</code>.</li>
</ul>

<p>Return the <strong>minimum</strong> cost to make <code>s</code> equal to <code>target</code>. If it&#39;s not possible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">target = &quot;abcdef&quot;, words = [&quot;abdef&quot;,&quot;abc&quot;,&quot;d&quot;,&quot;def&quot;,&quot;ef&quot;], costs = [100,1,1,10,5]</span></p>

<p><strong>Output:</strong> <span class="example-io">7</span></p>

<p><strong>Explanation:</strong></p>

<p>The minimum cost can be achieved by performing the following operations:</p>

<ul>
	<li>Select index 1 and append <code>&quot;abc&quot;</code> to <code>s</code> at a cost of 1, resulting in <code>s = &quot;abc&quot;</code>.</li>
	<li>Select index 2 and append <code>&quot;d&quot;</code> to <code>s</code> at a cost of 1, resulting in <code>s = &quot;abcd&quot;</code>.</li>
	<li>Select index 4 and append <code>&quot;ef&quot;</code> to <code>s</code> at a cost of 5, resulting in <code>s = &quot;abcdef&quot;</code>.</li>
</ul>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">target = &quot;aaaa&quot;, words = [&quot;z&quot;,&quot;zz&quot;,&quot;zzz&quot;], costs = [1,10,100]</span></p>

<p><strong>Output:</strong> <span class="example-io">-1</span></p>

<p><strong>Explanation:</strong></p>

<p>It is impossible to make <code>s</code> equal to <code>target</code>, so we return -1.</p>
</div>

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

<ul>
	<li><code>1 &lt;= target.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words.length == costs.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= target.length</code></li>
	<li>The total sum of <code>words[i].length</code> is less than or equal to <code>5 * 10<sup>4</sup></code>.</li>
	<li><code>target</code> and <code>words[i]</code> consist only of lowercase English letters.</li>
	<li><code>1 &lt;= costs[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class TrieNode:
    def __init__(self):
        self.suffix: Optional[TrieNode] = None
        self.dict: Optional[TrieNode] = None
        self.child = defaultdict(TrieNode)
        self.word_id = -1


class Trie:
    def __init__(self, words, costs):
        self.count = 0
        self.root = TrieNode()
        self.start: TrieNode = None
        self.root.suffix = self.root.dict = self.root

        for i, word in enumerate(words):
            node = self.root
            for letter in word:
                node = node.child[letter]
            if node.word_id == -1 or costs[i] < costs[node.word_id]:
                node.word_id = i

        queue = deque([self.root])
        while queue:
            node = queue.popleft()
            for letter, child in node.child.items():
                p = node.suffix

                while p != self.root and letter not in p.child:
                    p = p.suffix

                if node != self.root and letter in p.child:
                    child.suffix = p.child[letter]
                else:
                    child.suffix = self.root

                child.dict = child.suffix if child.suffix.word_id != -1 else child.suffix.dict
                queue.append(child)

        self.current: TrieNode = self.root

    def append(self, letter):
        while self.current != self.root and letter not in self.current.child:
            self.current = self.current.suffix

        if letter in self.current.child:
            self.current = self.current.child[letter]
            self.start = self.current if self.current.word_id != -1 else self.current.dict
        else:
            self.start = self.root


class Solution:
    def minimumCost(self, target: str, words: List[str], costs: List[int]) -> int:
        trie = Trie(words, costs)
        dp = [inf] * (len(target) + 1)
        dp[0] = 0

        for i in range(1, len(target) + 1):
            trie.append(target[i - 1])
            node = trie.start

            while node != trie.root:
                dp[i] = min(dp[i], dp[i - len(words[node.word_id])] + costs[node.word_id])
                node = node.dict

        return dp[len(target)] if dp[len(target)] != inf else -1
42900