<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) {
        
        int total=0;
        //Atleast(k)-Atleast(k+1)
        unordered_map<int ,int>freq;

        int start=0,end=0,n=nums.size(),count=0;

        //count of different integer in that window

        while(end<n)
        {
            freq[nums[end]]++; //add the freq of that number

            if(freq[nums[end]]==1)
            count++;

            //increase the window size tiil
            //number of differnt integer equal to k

            //Decrease the window size
            while(count==k)
            {
                total+=n-end;
                freq[nums[start]]--;
                if(freq[nums[start]]==0)
                count--;
                
                start++;
            }
            end++;

        }

        start=0,end=0,count=0;
        freq.clear();
        k++;

        
        while(end<n)
        {
            freq[nums[end]]++; //add the freq of that number

            if(freq[nums[end]]==1)
            count++;

            //increase the window size tiil
            //number of differnt integer equal to k

            //Decrease the window size
            while(count==k)
            {
                total-=n-end;
                freq[nums[start]]--;
                if(freq[nums[start]]==0)
                count--;
                
                start++;
            }
            end++;

        }
        return total;

    }
};
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