<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 atMostKDistinct(vector<int> &nums, int k){
        int count = 0; //stores no. of subarrays with <= k distinct nos.

        unordered_map<int, int> freq; //stores {num, freq}
        int i = 0; //farthest left edge possible for curr window
        for (int j = 0; j < nums.size(); ++j){ //right edge of curr window
            ++freq[nums[j]]; //updating freq

            while (freq.size() > k){ //updating i to make valid window
                --freq[nums[i]];
                if (freq[nums[i]] == 0){ freq.erase(nums[i]); }
                ++i;
            }

            count += j - i + 1; //all possible left edges for right edge j
        }

        return count;
    }

    int subarraysWithKDistinct(vector<int> &nums, int k){
        //return atMostKDistinct(nums, k) - atMostKDistinct(nums, k - 1);

        int count = 0;
        unordered_map<int, int> freq; //stores {num, freq}
        int j = 0; //curr right edge for curr window
        int right = 0; //farthest right edge possible for curr window
        for (int i = 0; i < nums.size(); ++i){
            while (j < nums.size() && freq.size() < k){ //updating i to make valid window
                ++freq[nums[j]];
                ++j;
            }

            if (freq.size() < k){ break; }

            while (right < nums.size() && freq.find(nums[right]) != freq.end()){ ++right; }

            count += right - (j - 1);

            if (--freq[nums[i]] == 0){ freq.erase(nums[i]); } //updating freq
        }
        return count;
    }
};
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