<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>

from typing import List

class Solution:
    def minValidStrings(self, words: List[str], target: str) -> int:
        n = len(target)
        inf = float('inf')
        dp = [inf] * (n + 1)
        dp[0] = 0  # Base case: empty string requires 0 prefixes

        # Initialize the Trie with list-based children
        # Each node has 26 children corresponding to 'a' to 'z'
        children = [[-1] * 26]
        is_end = [False]

        # Function to insert a word into the Trie
        for word in words:
            node = 0
            for char in word:
                idx = ord(char) - ord('a')
                if children[node][idx] == -1:
                    children[node][idx] = len(children)
                    children.append([-1] * 26)
                    is_end.append(False)
                node = children[node][idx]
                is_end[node] = True  # Mark the end of a prefix

        # Precompute the target characters as integers for faster access
        target_chars = [ord(c) - ord('a') for c in target]

        # Traverse the target string
        for i in range(n):
            if dp[i] == inf:
                continue
            node = 0
            j = i
            while j < n:
                idx = target_chars[j]
                if children[node][idx] == -1:
                    break  # No further prefix matches
                node = children[node][idx]
                j += 1
                if is_end[node]:
                    if dp[j] > dp[i] + 1:
                        dp[j] = dp[i] + 1

        return dp[n] if dp[n] != inf else -1

36594