<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>

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,mmx,sse2,sse3,sse4")
static const auto fast_io = []() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    return 0;
}();

class Solution {
public:
    int subarraysWithKDistinct(vector<int>& a, int k) {
        const int N = a.size();
        short counts[20001] = {0};
        int unique = 0;
        int res = 0;
        int l = 0, r = 0;
        for (int i = 0; i < N; ++i) {
            unique += (counts[a[i]] == 0);
            ++counts[a[i]];
            if (unique >= k) {
                if (unique > k) {
                    counts[a[r]] = 0;
                    --unique;
                    ++r;
                    l = r;
                }
                for (; r < i && counts[a[r]] > 1; r++) {
                    --counts[a[r]];
                }
                res += r - l + 1; 
            }
        }
        return res;
    }
};
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