<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 counting_sort():
            output = []
            counts = defaultdict(int)
            mini = nums[0]
            maxi = nums[0]
            for val in nums:
                counts[val] += 1
                mini = min(mini, val)
                maxi = max(maxi, val)
            
            for i in range(mini, maxi + 1):
                while counts[i]:
                    output.append(i)
                    counts[i] -= 1
            
            return output


        # def insertion_sort():
        #     for i in range(1, len(nums)):
        #         j = i
        #         while j > 0 and nums[j - 1] > nums[j]:
        #             nums[j], nums[j-1] = nums[j-1], nums[j]
        #             j -= 1
        #     return nums
        
        return counting_sort()
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