<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>

import math
class Solution:
    def sortArray(self, nums: List[int]) -> List[int]:
        M = max(nums)
        m = min(nums)
        if M == m:
            return nums
        normalized_lst = [(num-m)/(M-m) for num in nums]
        cells = [[] for num in nums]
        for num in normalized_lst:
            if num == 1:
                cells[-1].append(num)
            else:
                cells[math.floor(len(nums)*num)].append(num)
        for cell in cells:
            cell.sort()
        res = []
        for cell in cells:
            res+=cell
        for i in range(len(res)):
            res[i] = round(res[i]*(M-m)+m)
        return res

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