Count the Number of Good Subarrays
<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 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>
Array; Hash Table; Sliding Window
class Solution {
    public long countGood(int[] nums, int k) {
        Map<Integer, Integer> cnt = new HashMap<>();
        long ans = 0, cur = 0;
        int i = 0;
        for (int x : nums) {
            cur += cnt.merge(x, 1, Integer::sum) - 1;
            while (cur - cnt.get(nums[i]) + 1 >= k) {
                cur -= cnt.merge(nums[i++], -1, Integer::sum);
            }
            if (cur >= k) {
                ans += i + 1;
            }
        }
        return ans;
    }
}