<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:
        # finding window having k elems
        # then follow following algo
        # total_subarray_including_window = element_on_left*elements_on_right
        #
        arr = []
        max_num = max(nums)
        for i in range(len(nums)):
            if nums[i] == max_num:
                arr.append(i)
        if len(arr) < k:
            return 0
        lip = -1
        count = 0
       
        for i in range(len(arr) - k + 1):
            li = arr[i]
            ri = arr[i + k - 1]
            subs = (li - lip) * (len(nums) - ri)
            count += subs
            lip = li
            
        return count
829