<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]:
        nums = [[num] for num in nums]
        def mergeSort(nums):
            while len(nums)>1:
                new = []
                for i in range(0,len(nums),2):
                    if i==len(nums)-1:
                        new.append(nums[i])
                        continue
                    new.append(merge(nums[i], nums[i+1]))
                nums = new
            return nums[0]

        def merge(list1,list2):
            l = []
            while(len(list1)>0 and len(list2)>0):
                if list1[0]<=list2[0]:
                    l.append(list1.pop(0))
                else:
                    l.append(list2.pop(0))
            l.extend(list1)
            l.extend(list2)
            return l
        return mergeSort(nums)
            
        
        
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