<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, A: List[int]) -> bool:
        total = sum(A)
        # The ans does not change if we scale or shift the array
        for i in range(len(A)): A[i] = len(A) * A[i] - total
        # The above ensures that sum(A)=Avg(A)=avg(B)=avg(C)=sum(B)=sum(C)=0
        # So we are looking for a non-empty strict subset of A that sum to 0
        
        A.sort() #help prune the search a bit
        X = set()
        for a in A[:-1]: #excluding last element so it looks for STRICT subset
            X |= { a } | { a + x for x in X if x < 0}
            if 0 in X: return True
        return False
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