<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>

class Solution:
    @staticmethod
    def _merge(a: List[int], b: List[int]) -> List[int]:
        res = [0] * (len(a) + len(b))
        i, j, k = 0, 0, 0
        while k < len(res):
            if i < len(a) and (j >= len(b) or a[i] < b[j]):
                res[k] = a[i]
                i += 1
            else:
                res[k] = b[j]
                j += 1
            k += 1
        return res

    def sortArray(self, nums: List[int]) -> List[int]:
        chunks = [[n] for n in nums]
        while len(chunks) > 1:
            new_chunks = []
            for i in range(0, len(chunks), 2):
                if i == len(chunks) - 1:
                    new_chunks.append(chunks[i])
                else:
                    new_chunks.append(Solution._merge(chunks[i], chunks[i + 1]))
            chunks = new_chunks
        return chunks[0]

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