<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]:
        maxi, mini = float(-inf), float(inf)
        for val in set(nums):
            if val > maxi:
                maxi = val
            if val < mini:
                mini = val

        pos_arr = [0]*(maxi+1)
        neg_arr = [0]*(-1*mini)

        for val in nums:
            if val > -1:
                pos_arr[val] += 1
            else:
                neg_arr[val] += 1

        f = []
        n = len(neg_arr)
        for idx, val in enumerate(neg_arr):
            f.extend([-1 * (n-idx)]*val)

        for idx, val in enumerate(pos_arr):
            f.extend([idx]*val)
        
        return f
        
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