sort-an-array

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Description

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 necessairly unique.

 

Constraints:

• 1 <= nums.length <= 5 * 104
• -5 * 104 <= nums[i] <= 5 * 104

Solution(Python)

class Solution:
    def sortArray(self, nums: List[int]) -> List[int]:
        return self.countingSort(nums)

    # merge two sorted arrays
    # O(n) 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