Count Complete Subarrays in an Array
<p>You are given an array <code>nums</code> consisting of <strong>positive</strong> integers.</p>

<p>We call a subarray of an array <strong>complete</strong> if the following condition is satisfied:</p>

<ul>
	<li>The number of <strong>distinct</strong> elements in the subarray is equal to the number of distinct elements in the whole array.</li>
</ul>

<p>Return <em>the number of <strong>complete</strong> subarrays</em>.</p>

<p>A <strong>subarray</strong> is a contiguous non-empty part of an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,1,2,2]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The complete subarrays are the following: [1,3,1,2], [1,3,1,2,2], [3,1,2] and [3,1,2,2].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,5,5,5]
<strong>Output:</strong> 10
<strong>Explanation:</strong> The array consists only of the integer 5, so any subarray is complete. The number of subarrays that we can choose is 10.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 2000</code></li>
</ul>
Array; Hash Table; Sliding Window
class Solution:
    def countCompleteSubarrays(self, nums: List[int]) -> int:
        cnt = len(set(nums))
        ans, n = 0, len(nums)
        for i in range(n):
            s = set()
            for x in nums[i:]:
                s.add(x)
                if len(s) == cnt:
                    ans += 1
        return ans
