<p>Given an array of integers <code>nums</code>, sort the array in ascending order and return it.</p>

<p>You must solve the problem <strong>without using any built-in</strong> functions in <code>O(nlog(n))</code> time complexity and with the smallest space complexity possible.</p>

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

<pre>
<strong>Input:</strong> nums = [5,2,3,1]
<strong>Output:</strong> [1,2,3,5]
<strong>Explanation:</strong> After sorting the array, the positions of some numbers are not changed (for example, 2 and 3), while the positions of other numbers are changed (for example, 1 and 5).
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,1,1,2,0,0]
<strong>Output:</strong> [0,0,1,1,2,5]
<strong>Explanation:</strong> Note that the values of nums are not necessairly unique.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-5 * 10<sup>4</sup> &lt;= nums[i] &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

from collections import deque

class Solution:
    def sortArray(self, nums: List[int]) -> List[int]:
        def combineArrays(a, b):
            final = []
            aindex, bindex = 0, 0
            while 1:
                if aindex >= len(a) and bindex >= len(b):
                    break
                if bindex >= len(b) or (aindex < len(a) and a[aindex] < b[bindex]):
                    final.append(a[aindex])
                    aindex += 1
                else:
                    final.append(b[bindex])
                    bindex += 1
            return final
        
        n = [[x] for x in nums]
        nextN = []
        while len(n) > 1:
            for i in range(0, len(n), 2):
                a = n[i]
                if i + 1 < len(n):
                    b = n[i + 1]
                    nextN.append(combineArrays(a, b))
                else:
                    nextN.append(a)
            n = nextN
            nextN = []

        return n[0]
 
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