<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>3</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>3</sup></code></li>
	<li><code>target</code> consists only of lowercase English letters.</li>
</ul>

class TrieNode:
    def __init__(self):
        self.children = {}
        self.word = False
        self.startCounter = 0

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self,word):
        node = self.root
        for ch in word:
            if not ch in node.children:
                node.children[ch] = TrieNode()
            node = node.children[ch]
            node.startCounter += 1
        node.word = True
    
    @cache
    def startsWith(self,word):
        node = self.root
        for ch in word:
            if not ch in node.children:
                return False
            else:
                node = node.children[ch]
        return True

    def search(self, word):
        node = self.root
        for ch in word:
            if not ch in node.children:
                return False
            else:
                node = node.children[ch]
        return node.word

class Solution:
    def minValidStrings(self, words: List[str], target: str) -> int:
        

        T = Trie()
        for word in words:
            T.insert(word)

        N = len(target)
        @cache
        def dfs(i):
            if i >= N:
                return 0
            #search if target[i:j] is in the Trie?
            #find the LAST value for each j(i) that is in the Trie
            j = i+1
            
            #rewrite this so that it doesnt repeat the stepping in
            ##while j <= N and T.startsWith(target[i:j]):
            ##    j += 1

            j = i
            node = T.root
            while j < N and (ch:=target[j]) in node.children:
                node = node.children[ch]
                #print(ch, j, node.children.keys())
                j += 1

            #print(i,j)

            ####now we have this max j
            minVal = inf
            for k in range(i+1,j+1):
                minVal = min(minVal, dfs(k) + 1)

            return minVal
            #return dfs(j+1) + 1

        out = dfs(0)
        return out if isfinite(out) else -1

    #Trie solution with startsWith running via DP thorugh target
    #and then returning the minCount of this


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