<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the number of <strong>good</strong> subarrays of</em> <code>nums</code>.</p>

<p>A subarray <code>arr</code> is <strong>good</strong> if it there are <strong>at least </strong><code>k</code> pairs of indices <code>(i, j)</code> such that <code>i &lt; j</code> and <code>arr[i] == arr[j]</code>.</p>

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

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

<pre>
<strong>Input:</strong> nums = [1,1,1,1,1], k = 10
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only good subarray is the array nums itself.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,1,4,3,2,2,4], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are 4 different good subarrays:
- [3,1,4,3,2,2] that has 2 pairs.
- [3,1,4,3,2,2,4] that has 3 pairs.
- [1,4,3,2,2,4] that has 2 pairs.
- [4,3,2,2,4] that has 2 pairs.
</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], k &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def countGood(self, nums: List[int], k: int) -> int:
        cache = {}
        l, r = 0, 0
        res = 0
        curr_pairs = 0

        while l <= r:
            l_num = nums[l]
            
            if r < len(nums):
                r_num = nums[r]

            while curr_pairs >= k:
                l_num = nums[l]
                res += len(nums) + 1 - r
                curr_pairs -= (cache[l_num] - 1)
                cache[l_num] -= 1
                l += 1
            
            if r >= len(nums):
                return res
                
            if r_num in cache:
                curr_pairs += cache[r_num]
            cache[r_num] = cache.get(r_num, 0) + 1
            
            r += 1




        
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