<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>& nums, int k) {
        unordered_map<int, int> map;
        int left = 0;
        int len = 0;
        int result = 0;
        for (int i = 0; i < nums.size(); ++i) {
            ++map[nums[i]];

            if (map.size() > k) {
                int count = --map[nums[left]];
                if (count == 0) {
                    map.erase(nums[left]);
                }

                ++left;

                // reset len because we cannot have new element at nums[i]
                // with previous subarray len(s) with nums[left] since
                // it will exceed k
                len = 0;
            }

            // left marks left window of valid subarray
            // are there duplicates in the window of valid subarray?
            if (map.size() == k) {
                while (map[nums[left]] >= 2) {
                    --map[nums[left]];
                    ++left;
                    ++len;
                }

                // after the above loop, we guarantee that the element
                // at left is unique

                // the element at left is as far as we can count the number
                // of valid subarrays with k distinct elements for len
                
                result += (len + 1);
            }
        }

        return result;
    }
};
47044