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
func splitArraySameAverage(nums []int) bool {
	n := len(nums)
	if n == 1 {
		return false
	}
	s := 0
	for _, v := range nums {
		s += v
	}
	for i, v := range nums {
		nums[i] = v*n - s
	}
	m := n >> 1
	vis := map[int]bool{}
	for i := 1; i < 1<<m; i++ {
		t := 0
		for j, v := range nums[:m] {
			if (i >> j & 1) == 1 {
				t += v
			}
		}
		if t == 0 {
			return true
		}
		vis[t] = true
	}
	for i := 1; i < 1<<(n-m); i++ {
		t := 0
		for j, v := range nums[m:] {
			if (i >> j & 1) == 1 {
				t += v
			}
		}
		if t == 0 || (i != (1<<(n-m))-1 && vis[-t]) {
			return true
		}
	}
	return false
}