280. Wiggle Sort - Explanation

Description

Given an integer array nums, reorder it such that nums[0] <= nums[1] >= nums[2] <= nums[3]....

You may assume the input array always has a valid answer.

Example 1:

Input: nums = [3,5,2,1,6,4]

Output: [3,5,1,6,2,4]

Explanation: [1,6,2,5,3,4] is also accepted.

Example 2:

Input: nums = [6,6,5,6,3,8]

Output: [6,6,5,6,3,8]

Constraints:

1 <= nums.length <= 5 * 10^4
0 <= nums[i] <= 10^4
It is guaranteed that there will be an answer for the given input nums.

Follow Up: Could you solve the problem in O(n) time complexity?

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1. Max-Heap

class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """

        maxHeap = []
        for num in nums:
            heapq.heappush(maxHeap, -num)

        n = len(nums)
        for i in range(1, n, 2):
            nums[i] = -heapq.heappop(maxHeap)
        for i in range(0, n, 2):
            nums[i] = -heapq.heappop(maxHeap)

Time & Space Complexity

Time complexity: $O(n \log n)$
Space complexity: $O(n)$

2. Sorting

class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """

        nums.sort()
        for i in range(1, len(nums) - 1, 2):
            nums[i], nums[i + 1] = nums[i + 1], nums[i]

Time & Space Complexity

Time complexity: $O(n \log n)$
Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm.

3. Greedy - I

class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """

        for i in range(1, len(nums)):
            if ((i % 2 == 1 and nums[i] < nums[i - 1]) or
                (i % 2 == 0 and nums[i] > nums[i - 1])
            ):
                nums[i], nums[i - 1] = nums[i - 1], nums[i]

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

4. Greedy - II

class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """

        for i in range(1, len(nums)):
            if (i % 2) ^ (nums[i] > nums[i - 1]):
                nums[i], nums[i - 1] = nums[i - 1], nums[i]

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$