<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>

class Solution {
public:
    int subarraysWithKDistinct(vector<int>& a, int k) {
        int n = a.size();
        vector<int> b(n), c(n);
        int R = -1;
        vector<int> cnt(n + 1);
        int cur = 0;
        for (int L = 0; L < n; ++L) {
            while (R < n && cur < k) {
                ++R;
                if (R < n) {
                    cnt[a[R]]++;
                    if (cnt[a[R]] == 1) ++cur;
                }
            }
            b[L] = R;
            cnt[a[L]]--;
            if (cnt[a[L]] == 0) --cur;
        }
        ++k;
        fill(cnt.begin(), cnt.end(), 0);
        cur = 0;
        R = -1;
        for (int L = 0; L < n; ++L) {
            while (R < n && cur < k) {
                ++R;
                if (R < n) {
                    cnt[a[R]]++;
                    if (cnt[a[R]] == 1) ++cur;
                }
            }
            c[L] = R;
            cnt[a[L]]--;
            if (cnt[a[L]] == 0) --cur;
        }
        int ans = 0;
        for (int i = 0; i < n && b[i] < n; ++i) {
            ans += (c[i] - b[i]);
        }
        return ans;
    }
};
43581