Split Array With Same Average
<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>
Bit Manipulation; Array; Math; Dynamic Programming; Bitmask
class Solution:
    def splitArraySameAverage(self, nums: List[int]) -> bool:
        n = len(nums)
        if n == 1:
            return False
        s = sum(nums)
        for i, v in enumerate(nums):
            nums[i] = v * n - s
        m = n >> 1
        vis = set()
        for i in range(1, 1 << m):
            t = sum(v for j, v in enumerate(nums[:m]) if i >> j & 1)
            if t == 0:
                return True
            vis.add(t)
        for i in range(1, 1 << (n - m)):
            t = sum(v for j, v in enumerate(nums[m:]) if i >> j & 1)
            if t == 0 or (i != (1 << (n - m)) - 1 and -t in vis):
                return True
        return False
