<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:
    def sortArray(self, nums: List[int]) -> List[int]:

        def merge(left, mid, right):
            temp = []
            first = left
            second = mid + 1

            while first <= mid and second <= right:
                if nums[first] <= nums[second]:
                    temp.append(nums[first])
                    first += 1
                else:
                    temp.append(nums[second])
                    second += 1

            if first <= mid:
                while first <= mid:
                    temp.append(nums[first])
                    first += 1

            if second <= right:
                while second <= right:
                    temp.append(nums[second])
                    second += 1

            for i in range(left, right + 1):
                nums[i] = temp[i - left]

        def mergeSort(left, right):
            if left == right:
                return

            mid = (left + right) // 2

            mergeSort(left, mid)
            mergeSort(mid + 1, right)
            merge(left, mid, right)

        left, right = 0, len(nums) - 1
        mergeSort(left, right)

        return nums


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