<p>You are given an integer <code>n</code> representing the length of an unknown array that you are trying to recover. You are also given an array <code>sums</code> containing the values of all <code>2<sup>n</sup></code> <strong>subset sums</strong> of the unknown array (in no particular order).</p>

<p>Return <em>the array </em><code>ans</code><em> of length </em><code>n</code><em> representing the unknown array. If <strong>multiple</strong> answers exist, return <strong>any</strong> of them</em>.</p>

<p>An array <code>sub</code> is a <strong>subset</strong> of an array <code>arr</code> if <code>sub</code> can be obtained from <code>arr</code> by deleting some (possibly zero or all) elements of <code>arr</code>. The sum of the elements in <code>sub</code> is one possible <strong>subset sum</strong> of <code>arr</code>. The sum of an empty array is considered to be <code>0</code>.</p>

<p><strong>Note:</strong> Test cases are generated such that there will <strong>always</strong> be at least one correct answer.</p>

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

<pre>
<strong>Input:</strong> n = 3, sums = [-3,-2,-1,0,0,1,2,3]
<strong>Output:</strong> [1,2,-3]
<strong>Explanation: </strong>[1,2,-3] is able to achieve the given subset sums:
- []: sum is 0
- [1]: sum is 1
- [2]: sum is 2
- [1,2]: sum is 3
- [-3]: sum is -3
- [1,-3]: sum is -2
- [2,-3]: sum is -1
- [1,2,-3]: sum is 0
Note that any permutation of [1,2,-3] and also any permutation of [-1,-2,3] will also be accepted.
</pre>

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

<pre>
<strong>Input:</strong> n = 2, sums = [0,0,0,0]
<strong>Output:</strong> [0,0]
<strong>Explanation:</strong> The only correct answer is [0,0].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 4, sums = [0,0,5,5,4,-1,4,9,9,-1,4,3,4,8,3,8]
<strong>Output:</strong> [0,-1,4,5]
<strong>Explanation:</strong> [0,-1,4,5] is able to achieve the given subset sums.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 15</code></li>
	<li><code>sums.length == 2<sup>n</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= sums[i] &lt;= 10<sup>4</sup></code></li>
</ul>



class Solution:
    def __init__(self):
        self.ans = []
    
    def findX(self, array):
        n = len(array)
        if n <= 1 or n % 2 == 1:
            return
        
        if n == 2:
            self.ans.append(array[0] + array[1])
            return
        
        p = 1
        subhaszero = False
        while p < n:
            x = array[p] - array[0]
            subarray, otherarray = [], []
            q = deque(array)
            freq = Counter(array)
            
            while q:
                v = q.popleft()
                if freq[v] > 0:
                    freq[v] -= 1
                    v += x
                    if freq[v] > 0:
                        freq[v] -= 1
                        subarray.append(v - x)
                        subhaszero |= (v == x)
                        otherarray.append(v)
                    else:
                        break
            
            if len(subarray) == n // 2:
                if subhaszero:
                    self.ans.append(x)
                    self.findX(subarray)
                else:
                    self.ans.append(-x)
                    self.findX(otherarray)
                return
            
            p += 1
            while p < n and array[p] == array[p - 1]:
                p += 1
                
        return
    
    def recoverArray(self, n, sums):
        self.ans.clear()
        sums.sort()
        self.findX(sums)
        return self.ans if len(self.ans) == n else []

1224