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 {
    public boolean splitArraySameAverage(int[] nums) {
        int n = nums.length;
        if (n == 1) {
            return false;
        }
        int s = Arrays.stream(nums).sum();
        for (int i = 0; i < n; ++i) {
            nums[i] = nums[i] * n - s;
        }
        int m = n >> 1;
        Set<Integer> vis = new HashSet<>();
        for (int i = 1; i < 1 << m; ++i) {
            int t = 0;
            for (int j = 0; j < m; ++j) {
                if (((i >> j) & 1) == 1) {
                    t += nums[j];
                }
            }
            if (t == 0) {
                return true;
            }
            vis.add(t);
        }
        for (int i = 1; i < 1 << (n - m); ++i) {
            int t = 0;
            for (int j = 0; j < (n - m); ++j) {
                if (((i >> j) & 1) == 1) {
                    t += nums[m + j];
                }
            }
            if (t == 0 || (i != (1 << (n - m)) - 1) && vis.contains(-t)) {
                return true;
            }
        }
        return false;
    }
}