<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 sum1 , sum2 , i , j;
        sum1 = sum2 = i = j = 0;

        unordered_map<int , int>map;

        while(j < nums.size())
        {
            map[nums[j]]++;

            while(map.size() > k and i < j)
            {
                map[nums[i]]--;
                if (map[nums[i]] == 0) 
                {
                    map.erase(nums[i]);
                }
                i++;
                
            }
            if(map.size() <= k)
            {
                sum1 += j - i + 1;
                cout<<sum1<<endl;
            }
            j++;
        }

        --k;

        j = 0;
        i = 0;

        map.clear();

        while(j < nums.size())
        {
            map[nums[j]]++;

            while(map.size() > k and i < j)
            {
                map[nums[i]]--;
                if (map[nums[i]] == 0) 
                {
                    map.erase(nums[i]);
                }
                i++;
                
            }
            if(map.size() <= k)
            {
                sum2 += j - i + 1;
            }
            j++;
        }

        // cout<<sum1 <<" "<<sum2<<endl;
        return sum1 - sum2;
        
    }
};
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