<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>& v, int k) {
        priority_queue<int,vector<int>, greater<int>>pq;
        map<int,int>mp;
        map<int,int>max_index;
        int n=v.size(), l=0, ans=0;
        for(int i=0;i<n;i++)
        {
            mp[v[i]]++;
            max_index[v[i]]=i;
            pq.push(i);
            if(mp.size()>=k){
                while(mp.size()>k)
                {
                    mp[v[l]]--;
                    if(mp[v[l]]==0)
                    {
                        mp.erase(v[l]);
                    }
                    l++;
                }
                int min_of_max_indexes = i;
                while(!pq.empty())
                {
                    int idx = pq.top();
                    
                    if(max_index[v[idx]] == idx && mp.find(v[idx])!=mp.end()){
                        min_of_max_indexes = idx;
                        break;
                    }
                    else{
                        pq.pop();
                    }
                }
                ans+=(min_of_max_indexes-l+1);
                cout<<i<<" "<<l<<" "<<min_of_max_indexes<<" "<<ans<<endl;
                
            }
        }
        return ans;

        
    }
};
// [2,2,1,2,2,2,1,1]
59509