<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>

from typing import List

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        # find max element
        maxNum = nums[0]
        for num in nums:
            maxNum = max(maxNum, num)

        # use sliding window approach to determine number of subarrays
        left, right = 0, -1
        maxCount = 0
        out = 0
        while right < len(nums) and left - 1 <= right:
            if maxCount == k:
                while left - 1 <= right and nums[left] != maxNum:
                    out += len(nums) - right
                    left += 1
                out += len(nums) - right
                left += 1
                maxCount -= 1
            elif right + 1 < len(nums) and maxCount < k:
                right += 1
                if nums[right] == maxNum:
                    maxCount += 1
            else:
                left += 1
        
        return out
    
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