<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>

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        maximum = max(nums)
        indices_of_maxima = [i for i, x in enumerate(nums) if x == maximum]

        if len(indices_of_maxima) < k:
            return 0
        if len(indices_of_maxima) == k:
            return (indices_of_maxima[0] + 1) * (len(nums) - indices_of_maxima[-1])

        result = 0
        n = len(nums)
        indices_of_maxima = [-1] + indices_of_maxima
        for idx_1, idx_2, idx_3 in zip(
            indices_of_maxima[:-k],
            indices_of_maxima[1 : max(-k + 1, n)],
            indices_of_maxima[k:],
        ):
            result += (idx_2 - idx_1) * (n - idx_3)

        return result

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