<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 Node:
    def __init__(self):
        self.children = {}
        self.fail = None
        self.output = []

from collections import deque

class Solution:
    def minimumCost(self, target: str, words: List[str], costs: List[int]) -> int:
        # Build the Aho-Corasick automaton
        root = Node()
        for word, cost in zip(words, costs):
            node = root
            for c in word:
                if c not in node.children:
                    node.children[c] = Node()
                node = node.children[c]
            # Store minimal cost for identical words
            found = False
            for idx, (w_len, existing_cost) in enumerate(node.output):
                if w_len == len(word):
                    if cost < existing_cost:
                        node.output[idx] = (len(word), cost)
                    found = True
                    break
            if not found:
                node.output.append((len(word), cost))

        # Build failure links
        queue = deque()
        for child in root.children.values():
            child.fail = root
            queue.append(child)

        while queue:
            current_node = queue.popleft()
            for char, child_node in current_node.children.items():
                fail_node = current_node.fail
                while fail_node is not None and char not in fail_node.children:
                    fail_node = fail_node.fail
                child_node.fail = fail_node.children[char] if fail_node and char in fail_node.children else root
                # Merge output
                child_node.output += child_node.fail.output if child_node.fail else []
                queue.append(child_node)

        n = len(target)
        dp = [float('inf')] * (n + 1)
        dp[0] = 0

        node = root
        for i in range(n):
            c = target[i]
            while node is not None and c not in node.children:
                node = node.fail
            node = node.children[c] if node and c in node.children else root
            # For all patterns that end here, update dp
            for length, cost in node.output:
                start_pos = i - length + 1
                if dp[start_pos] != float('inf'):
                    dp[i + 1] = min(dp[i + 1], dp[start_pos] + cost)

        return dp[n] if dp[n] != float('inf') else -1

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