<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 Solution:
    def minValidStrings(self, words: List[str], target: str) -> int:
        p = 53 # 31, 53, 29791, 11111, 111111
        M = 10**9 + 7
        power = [1]

        n = len(target)
        for _ in range(n + 1):
            power.append((power[-1] * p) % M)

        # compute the hash for all prefix
        prefixes = set()
        for w in words:
            m = len(w)
            h = 0
            for c in w:
                h = (h * p + (ord(c) - 96)) % M
                prefixes.add(h)
        

        """
        Important observation:
        
        - If longest match at dp[r] starts at l 
        then longest match at dp[r + 1] cannot starts before l

        => Thus a sliding window is applicable here.

        - Additionally, the dp array will consists of spans of equal numbers, i.e. 1112222223444...
        => So we can just keep track when l move to a new span, no need to store the whole dp array.
        """

        h = 0
        l = 0
        span_start = 0
        ans = 1
        for r in range(n):
            # expand the window one step to the right
            h = (h * p + (ord(target[r]) - 96)) % M

            # shrink the window from the left until found a valid prefix
            while l <= r and h not in prefixes:
                x = ord(target[l]) - 96
                h = (h - (x * power[r - l])) % M
                l += 1

            # early stop if no valid prefix found
            if l > r:
                return -1

            # when l moves to a new span is also when r starts the next span
            if l > span_start:
                ans += 1
                span_start = r
        
        return ans
24471