<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:
        good_sets = set()
        counts = {} # count
        good_pairs = 0
        left,right = 0,0 
        res = 0
        for right in range(len(nums)):
            # build map
            val = nums[right]
            if val not in counts:
                counts[val] = 1
            else:
                counts[val]+=1

            # take previous counts +  
            if counts[val]>=1:
                good_pairs = good_pairs+ (counts[val]-1)

            # if i reach my good sets then add the good_sets
            if good_pairs >=k:
                good_sets.add((left,right))
                

            # if my good_pairs are greater than k i can reduce it down
            while good_pairs >=k:

                counts[nums[left]]-=1

                # if count of left value is 2 then removing that will remove a single good pair
                if counts[ nums[left] ] >= 1:
                    good_pairs = good_pairs - (counts [nums[left] ])
                
                left+=1

                if good_pairs>=k:
                    good_sets.add((left,right))

            res +=left
            
        return res
612