<p>You are given an integer array <code>nums</code>.</p>

<p>You should move each element of <code>nums</code> into one of the two arrays <code>A</code> and <code>B</code> such that <code>A</code> and <code>B</code> are non-empty, and <code>average(A) == average(B)</code>.</p>

<p>Return <code>true</code> if it is possible to achieve that and <code>false</code> otherwise.</p>

<p><strong>Note</strong> that for an array <code>arr</code>, <code>average(arr)</code> is the sum of all the elements of <code>arr</code> over the length of <code>arr</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6,7,8]
<strong>Output:</strong> true
<strong>Explanation:</strong> We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have an average of 4.5.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,1]
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 30</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def splitArraySameAverage(self, nums: List[int]) -> bool:
        n = len(nums)
        a = sum(nums)
        s = [e*n-a for e in nums]
        if len(nums) < 2:
            return False
        if 0 in s:
            return True
        neg = [-e for e in s if e < 0]
        pos = [e for e in s if e > 0]
        sneg = set({0})
        for n in neg:
            sneg = sneg.union({n + e for e in sneg})
        spos = set({0})
        for p in pos:
            spos = spos.union({p + e for e in spos})
        sneg.remove(0)
        spos.remove(0)
        sneg.remove(sum(neg))
        spos.remove(sum(pos))
        return len(sneg.intersection(spos)) > 0
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