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

<p>A <strong>good array</strong> is an array where the number of different integers in that array is exactly <code>k</code>.</p>

<ul>
	<li>For example, <code>[1,2,3,1,2]</code> has <code>3</code> different integers: <code>1</code>, <code>2</code>, and <code>3</code>.</li>
</ul>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,1,2,3], k = 2
<strong>Output:</strong> 7
<strong>Explanation:</strong> Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2]
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,1,3,4], k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i], k &lt;= nums.length</code></li>
</ul>
Array; Hash Table; Counting; Sliding Window
class Solution {
    public int subarraysWithKDistinct(int[] nums, int k) {
        int[] left = f(nums, k);
        int[] right = f(nums, k - 1);
        int ans = 0;
        for (int i = 0; i < nums.length; ++i) {
            ans += right[i] - left[i];
        }
        return ans;
    }

    private int[] f(int[] nums, int k) {
        int n = nums.length;
        int[] cnt = new int[n + 1];
        int[] pos = new int[n];
        int s = 0;
        for (int i = 0, j = 0; i < n; ++i) {
            if (++cnt[nums[i]] == 1) {
                ++s;
            }
            for (; s > k; ++j) {
                if (--cnt[nums[j]] == 0) {
                    --s;
                }
            }
            pos[i] = j;
        }
        return pos;
    }
}