<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(a, b):
            res = []
            l, r = 0, 0
            while l < len(a) and r < len(b):
                if a[l] < b[r]:
                    res.append(a[l])
                    l += 1
                else:
                    res.append(b[r])
                    r += 1
            res.extend(a[l:])
            res.extend(b[r:])
            return res
        
        def sort(arr):
            if len(arr) <= 1:
                return arr
            n = len(arr)
            return merge(sort(arr[:n//2]), sort(arr[n//2:]))
        
        return sort(nums)
    
    def sortArray(self, nums: List[int]) -> List[int]:
        
        cnt = [0]*1000001
        min_val = 5 * 10**4
        
        for n in nums:
            cnt[n + min_val] += 1
        
        p = 0
        for v, c in enumerate(cnt):
            for _ in range(c):
                nums[p] = v - min_val 
                p += 1
        return nums
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