Given an array of integers nums, sort the array in ascending order and return it.
You must solve the problem without using any built-in functions in O(nlog(n)) time complexity and with the smallest space complexity possible.
Example 1:
Input: nums = [5,2,3,1]
Output: [1,2,3,5]
Explanation: 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).
Example 2:
Input: nums = [5,1,1,2,0,0]
Output: [0,0,1,1,2,5]
Explanation: Note that the values of nums are not necessarily unique.
Constraints:
1 <= nums.length <= 5 * 104-5 * 104 <= nums[i] <= 5 * 104
## Solution(Python)
```Python
from random import randint
class Solution:
def sortArray(self, nums: List[int]) -> List[int]:
return self.quicksort(nums)
# merge two sorted arrays
# O(n LOGN) time complexity
# O(n) space complexity
def mergeSort(self, nums: List[int]) -> List[int]:
if len(nums) <= 1:
return nums
mid = len(nums) // 2
left = self.mergeSort(nums[:mid])
right = self.mergeSort(nums[mid:])
return self.merge(left, right)
# Heap sort
# O(nlogn) time complexity
# O(n) space complexity
def heapSort(self, nums: List[int]) -> List[int]:
heapq.heapify(nums)
return [heapq.heappop(nums) for _ in range(len(nums))]
# Counting Sort
# Time Complexity: O(n + k)
# SpaceComplexity : O(k)
def countingSort(self, nums: List[int]) -> List[int]:
min_num, max_num = min(nums), max(nums)
count = [0] * (max_num - min_num + 1)
for num in nums:
count[num - min_num] += 1
result = []
for i, c in enumerate(count):
result.extend([i + min_num] * c)
return result
def merge(self, left: List[int], right: List[int]) -> List[int]:
result = []
i = j = 0
while i < len(left) and j < len(right):
if left[i] < right[j]:
result.append(left[i])
i += 1
else:
result.append(right[j])
j += 1
result.extend(left[i:])
result.extend(right[j:])
return result
# Counting Sort
# Time Complexity: O(n ^ 2)
# SpaceComplexity : O(logn)
def quicksort(self, nums: List[int]) -> List[int]:
def quick_sort(left ,right):
if left >= right:
return
pivot = nums[randint(left,right)]
i = left -1
j = right + 1
k = left
while k < j:
if nums[k] < pivot:
i += 1
nums[i], nums[k] = nums[k], nums[i]
k += 1
elif nums[k] > pivot:
# Move element to the "greater than" region
j -= 1
nums[j], nums[k] = nums[k], nums[j]
# Don't increment k as we need to examine the swapped element
else:
# Element equals pivot, just move to next element
k += 1
# Recursively sort the "less than" and "greater than" regions
# Elements equal to pivot are already in correct position
quick_sort(left, i) # Sort elements < pivot
quick_sort(j, right) # Sort elements > pivot
# Sort the entire array in-place
quick_sort(0, len(nums) - 1)
return nums
```