<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) {
        map<int,int> mp,last;
        int ans=0,j=0;
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> pq;
        for(int i=0;i<nums.size();i++){
            mp[nums[i]]++;
            last[nums[i]]=i;
            pq.push({i,nums[i]});
            while(mp.size()>k){
                mp[nums[j]]--;
                if(mp[nums[j]]==0)
                    mp.erase(nums[j]);
                j++;
            }
            if(mp.size()==k){
                while(true){
                    int idx=pq.top().first;
                    int num=pq.top().second;
                    if(mp.find(num)==mp.end() || idx!=last[num])
                        pq.pop();
                    else{
                        // cout<<i<<" "<<idx<<" "<<j<<endl;
                        ans+=idx-j+1;
                        break;
                    }
                }
            }
            
        }
        return ans;
    }
};
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