<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) {
        // Array to store the count of distinct values encountered
        vector<int> distinctCount(nums.size() + 1, 0);

        int totalCount = 0;
        int left = 0;
        int right = 0;
        int currCount = 0;

        while (right < nums.size()) {
            // Increment the count of the current element in the window
            if (++distinctCount[nums[right++]] == 1) {
                // If encountering a new distinct element, decrement K
                k--;
            }

            // If K becomes negative, adjust the window from the left
            if (k < 0) {
                // Move the left pointer until the count of distinct elements
                // becomes valid again
                --distinctCount[nums[left++]];
                k++;
                currCount = 0;
            }

            // If K becomes zero, calculate subarrays
            if (k == 0) {
                // While the count of left remains greater than 1, keep
                // shrinking the window from the left
                while (distinctCount[nums[left]] > 1) {
                    --distinctCount[nums[left++]];
                    currCount++;
                }
                // Add the count of subarrays with k distinct elements to the
                // total count
                totalCount += (currCount + 1);
            }
        }
        return totalCount;
    }
};
41504