Count Substrings That Can Be Rearranged to Contain a String I
<p>You are given two strings <code>word1</code> and <code>word2</code>.</p>

<p>A string <code>x</code> is called <strong>valid</strong> if <code>x</code> can be rearranged to have <code>word2</code> as a <span data-keyword="string-prefix">prefix</span>.</p>

<p>Return the total number of <strong>valid</strong> <span data-keyword="substring-nonempty">substrings</span> of <code>word1</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">word1 = &quot;bcca&quot;, word2 = &quot;abc&quot;</span></p>

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

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

<p>The only valid substring is <code>&quot;bcca&quot;</code> which can be rearranged to <code>&quot;abcc&quot;</code> having <code>&quot;abc&quot;</code> as a prefix.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">word1 = &quot;abcabc&quot;, word2 = &quot;abc&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">10</span></p>

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

<p>All the substrings except substrings of size 1 and size 2 are valid.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">word1 = &quot;abcabc&quot;, word2 = &quot;aaabc&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">0</span></p>
</div>

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

<ul>
	<li><code>1 &lt;= word1.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= word2.length &lt;= 10<sup>4</sup></code></li>
	<li><code>word1</code> and <code>word2</code> consist only of lowercase English letters.</li>
</ul>
Hash Table; String; Sliding Window
class Solution:
    def validSubstringCount(self, word1: str, word2: str) -> int:
        if len(word1) < len(word2):
            return 0
        cnt = Counter(word2)
        need = len(cnt)
        ans = l = 0
        win = Counter()
        for c in word1:
            win[c] += 1
            if win[c] == cnt[c]:
                need -= 1
            while need == 0:
                if win[word1[l]] == cnt[word1[l]]:
                    need += 1
                win[word1[l]] -= 1
                l += 1
            ans += l
        return ans
