<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) {
        vector<int> c(20001);
        int n=nums.size();
        int i=0;
        int a=0;
        int ans=0;
        while(i<n){
            c[nums[i]]++;
            if(c[nums[i]]==1){
                a++;
            }
            if(a==k){
                break;
            }
            i++;
        }
        if(i==n){
            return 0;
        }
        int j=0;
        while(i<n){
            int i2=i;
            while(i2<n){
                ans++;
                i2++;
                if(i2>=n){
                    break;
                }
                if(c[nums[i2]]==0){
                    break;
                }
               // c[nums[i2]]++;
            }
            c[nums[j]]--;
            if(c[nums[j]]>0){
                j++;
            }
            else{
                j++;
                i++;
                while(i<n){
                    c[nums[i]]++;
                    if(c[nums[i]]==1){
                        break;
                    }
                    else{
                        i++;
                    }
                }

            }

        }
        return ans;
        
    }
};
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