<p>You are given an array of strings <code>words</code> and a string <code>target</code>.</p>

<p>A string <code>x</code> is called <strong>valid</strong> if <code>x</code> is a <span data-keyword="string-prefix">prefix</span> of <strong>any</strong> string in <code>words</code>.</p>

<p>Return the <strong>minimum</strong> number of <strong>valid</strong> strings that can be <em>concatenated</em> to form <code>target</code>. If it is <strong>not</strong> possible to form <code>target</code>, 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">words = [&quot;abc&quot;,&quot;aaaaa&quot;,&quot;bcdef&quot;], target = &quot;aabcdabc&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

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

<p>The target string can be formed by concatenating:</p>

<ul>
	<li>Prefix of length 2 of <code>words[1]</code>, i.e. <code>&quot;aa&quot;</code>.</li>
	<li>Prefix of length 3 of <code>words[2]</code>, i.e. <code>&quot;bcd&quot;</code>.</li>
	<li>Prefix of length 3 of <code>words[0]</code>, i.e. <code>&quot;abc&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">words = [&quot;abababab&quot;,&quot;ab&quot;], target = &quot;ababaababa&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

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

<p>The target string can be formed by concatenating:</p>

<ul>
	<li>Prefix of length 5 of <code>words[0]</code>, i.e. <code>&quot;ababa&quot;</code>.</li>
	<li>Prefix of length 5 of <code>words[0]</code>, i.e. <code>&quot;ababa&quot;</code>.</li>
</ul>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">words = [&quot;abcdef&quot;], target = &quot;xyz&quot;</span></p>

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

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

<ul>
	<li><code>1 &lt;= words.length &lt;= 100</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 5 * 10<sup>4</sup></code></li>
	<li>The input is generated such that <code>sum(words[i].length) &lt;= 10<sup>5</sup></code>.</li>
	<li><code>words[i]</code> consists only of lowercase English letters.</li>
	<li><code>1 &lt;= target.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>target</code> consists only of lowercase English letters.</li>
</ul>

class AhoCorasick:
    def __init__(self):
        self.root = { "suffix": None, "$": 0}
        
    def build(self, patterns: List[str]): 
        for pattern in patterns:
            node = self.root
            size = 0
            for ch in pattern:
                if ch not in node: node[ch] = {}
                node = node[ch]
                size += 1
                node["$"] = size
        queue = deque([self.root])
        while queue:
            for _ in range(len(queue)):
                node = queue.popleft()
                for ch, child in node.items():
                    if ch not in ( "suffix", "$"): 
                        suffix = node["suffix"]
                        while suffix and ch not in suffix: suffix = suffix["suffix"]
                        if suffix: child["suffix"] = suffix[ch]
                        else: child["suffix"] = self.root
                        queue.append(child)

    def match(self, text: str):
        ans = []
        node = self.root
        for ch in text:
            while ch not in node and node["suffix"]:
                node = node["suffix"]
            if ch in node: node = node[ch]
            ans.append(node["$"])
        return ans 
    

class Solution:
    def minValidStrings(self, words: List[str], target: str) -> int:
        n = len(target)
        trie = AhoCorasick()
        trie.build(words)
        dp = [inf]*(n+1)
        dp[n] = 0
        for i, x in reversed(list(enumerate(trie.match(target)))):
            if x == 0: return -1 
            dp[i+1-x] = min(dp[i+1-x], dp[i+1] + 1)
            print(f"i:{i} x:{x}")
        return dp[0] 
49633