<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:
        max_num = 0
        for num in nums:
            max_num = max(num, max_num) 
        
       
        new_nums = []
        for num in nums:
            if num == max_num:
                new_nums.append(1)
            else:
                new_nums.append(0)
        
        ans, count_less_than_k, start = 0, 0, 0
        N = len(new_nums)
        for end in range(N):
            count_less_than_k += new_nums[end]
            
            while count_less_than_k >= k:
                count_less_than_k -= new_nums[start]
                start += 1
            
            ans += end - start + 1
        
        # Calculate the total number of subarrays and subtract the count of invalid ones
        return N * (N + 1) // 2 - ans

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