<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(nums: List[int], left: int, mid: int, right: int) -> None:
            left_nums, right_nums = nums[left:mid+1], nums[mid+1:right+1]

            i, j, k = left, 0, 0
            while j < len(left_nums) and k < len(right_nums):
                if left_nums[j] < right_nums[k]:
                    nums[i] = left_nums[j]
                    j += 1
                else:
                    nums[i] = right_nums[k]
                    k += 1
                i += 1
            
            while j < len(left_nums):
                nums[i] = left_nums[j]
                j += 1
                i += 1

            while k < len(right_nums):
                nums[i] = right_nums[k]
                k += 1
                i += 1

            return

        def merge_sort(nums: List[int], left: int, right: int) -> None:
            """ In-place: bottom up """
            if left >= right:
                return

            # divide
            mid = left + (right - left) // 2
            merge_sort(nums, left, mid)
            merge_sort(nums, mid+1, right)

            # conquer
            merge(nums, left, mid, right)
            return

        merge_sort(nums, 0, len(nums)-1)
        return nums


        def partition(nums: List[int], left: int, right: int) -> None:
            # define the pivot value and index
            pivot_idx = right

            # partition: moving smaller elems to the left and bigger elems to the right
            # two pointers: windown[L+1:R+1] bigger values
            l, r = left - 1, left
            for r in range(left, right):
                if nums[r] < nums[pivot_idx]:
                    # find the left most bigger value
                    l += 1
                    # swap the foudn smaller value with the left most bigger value
                    tmp = nums[l]
                    nums[l] = nums[r]
                    nums[r] = nums[l]

            l += 1
            tmp = nums[pivot_idx]
            nums[pivot_idx] = nums[l]
            nums[l] = tmp
            
            return l
                    
        def quick_sort(nums: List[int], left: int, right: int) -> None:
            """ In-place sorting: top-down """
            if left >= right:
                return
            
            # conquer: perform partition around the pivot (i.e., the last one)
            pivot_idx = partition(nums, left, right)

            # divide: split the nums into twp part
            quick_sort(nums, left, pivot_idx - 1)
            quick_sort(nums, pivot_idx + 1, right)
            return 

        quick_sort(nums, 0, len(nums)-1)
        return nums

        def brute_force(nums: List[int]) -> List[int]:
            # loop from left -> right
            for i in range(0, len(nums)):
                # find the minimal number from i to len(nums) - 1
                min_idx = i
                for j in range(i+1, len(nums)):
                    if nums[j] < nums[min_idx]:
                        min_idx = j
                
                # swap the nums[i] with the minimal number
                tmp = nums[i]
                nums[i] = nums[min_idx]
                nums[min_idx] = tmp
            
            return nums

        return brute_force(nums)


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