<p>You are given a <strong>0-indexed </strong>integer array <code>nums</code>.</p>

<p>The <strong>distinct count</strong> of a subarray of <code>nums</code> is defined as:</p>

<ul>
	<li>Let <code>nums[i..j]</code> be a subarray of <code>nums</code> consisting of all the indices from <code>i</code> to <code>j</code> such that <code>0 &lt;= i &lt;= j &lt; nums.length</code>. Then the number of distinct values in <code>nums[i..j]</code> is called the distinct count of <code>nums[i..j]</code>.</li>
</ul>

<p>Return <em>the sum of the <strong>squares</strong> of <strong>distinct counts</strong> of all subarrays of </em><code>nums</code>.</p>

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

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

<pre>
<strong>Input:</strong> nums = [1,2,1]
<strong>Output:</strong> 15
<strong>Explanation:</strong> Six possible subarrays are:
[1]: 1 distinct value
[2]: 1 distinct value
[1]: 1 distinct value
[1,2]: 2 distinct values
[2,1]: 2 distinct values
[1,2,1]: 2 distinct values
The sum of the squares of the distinct counts in all subarrays is equal to 1<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup> + 2<sup>2</sup> + 2<sup>2</sup> + 2<sup>2</sup> = 15.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Three possible subarrays are:
[1]: 1 distinct value
[1]: 1 distinct value
[1,1]: 1 distinct value
The sum of the squares of the distinct counts in all subarrays is equal to 1<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup> = 3.</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution:
    def sumCounts(self, nums: List[int]) -> int:
        a = []
        for i in range(1,len(nums)+1):
            for j in range(len(nums)-i+1):
                sa = nums[j:j+i]
                ss = set(sa)
                a.append(ss)
        res = sum(len(i)**2 for i in a)

        return res
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