Count Substrings With K-Frequency Characters I
<p>Given a string <code>s</code> and an integer <code>k</code>, return the total number of <span data-keyword="substring-nonempty">substrings</span> of <code>s</code> where <strong>at least one</strong> character appears <strong>at least</strong> <code>k</code> times.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;abacb&quot;, k = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

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

<p>The valid substrings are:</p>

<ul>
	<li><code>&quot;aba&quot;</code> (character <code>&#39;a&#39;</code> appears 2 times).</li>
	<li><code>&quot;abac&quot;</code> (character <code>&#39;a&#39;</code> appears 2 times).</li>
	<li><code>&quot;abacb&quot;</code> (character <code>&#39;a&#39;</code> appears 2 times).</li>
	<li><code>&quot;bacb&quot;</code> (character <code>&#39;b&#39;</code> appears 2 times).</li>
</ul>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;abcde&quot;, k = 1</span></p>

<p><strong>Output:</strong> <span class="example-io">15</span></p>

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

<p>All substrings are valid because every character appears at least once.</p>
</div>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 3000</code></li>
	<li><code>1 &lt;= k &lt;= s.length</code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
</ul>
Hash Table; String; Sliding Window
function numberOfSubstrings(s: string, k: number): number {
    let [ans, l] = [0, 0];
    const cnt: number[] = Array(26).fill(0);
    for (const c of s) {
        const x = c.charCodeAt(0) - 'a'.charCodeAt(0);
        ++cnt[x];
        while (cnt[x] >= k) {
            --cnt[s[l++].charCodeAt(0) - 'a'.charCodeAt(0)];
        }
        ans += l;
    }
    return ans;
}
