class Solution: def sortArray(self, nums): def merge(left, right): i, j = 0, 0 result = [] 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 def merge_sort(nums): if len(nums) <= 1: return nums mid = len(nums) // 2 left = merge_sort(nums[:mid]) l right = merge_sort(nums[mid:]) return merge(left, right) return merge_sort(nums) give a detailed but 2-3 sentence analysis of the time and complexity and space complexity of the algorithm in big O notation
class Solution:
def sortArray(self, nums):
def merge(left, right):
i, j = 0, 0
result = []
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
def merge_sort(nums):
if len(nums) <= 1:
return nums
mid = len(nums) // 2
left = merge_sort(nums[:mid])
l
right = merge_sort(nums[mid:])
return merge(left, right)
return merge_sort(nums)
give a detailed but 2-3 sentence analysis of the time and complexity and space complexity of the algorithm in big O notation
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