Count Subarrays Where Max Element Appears at Least K Times
<p>You are given an integer array <code>nums</code> and a <strong>positive</strong> integer <code>k</code>.</p>

<p>Return <em>the number of subarrays where the <strong>maximum</strong> element of </em><code>nums</code><em> appears <strong>at least</strong> </em><code>k</code><em> times in that subarray.</em></p>

<p>A <strong>subarray</strong> is a contiguous sequence of elements within an array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,2,3,3], k = 2
<strong>Output:</strong> 6
<strong>Explanation:</strong> The subarrays that contain the element 3 at least 2 times are: [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,4,2,1], k = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> No subarray contains the element 4 at least 3 times.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>
Array; Sliding Window
function countSubarrays(nums: number[], k: number): number {
    const mx = Math.max(...nums);
    const n = nums.length;
    let [cnt, j] = [0, 0];
    let ans = 0;
    for (const x of nums) {
        for (; j < n && cnt < k; ++j) {
            cnt += nums[j] === mx ? 1 : 0;
        }
        if (cnt < k) {
            break;
        }
        ans += n - j + 1;
        cnt -= x === mx ? 1 : 0;
    }
    return ans;
}
