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Given an integer array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]... (also known as wiggle sort). You may assume all input arrays have any length and the rearrangement must be in-place (i.e., do not allocate extra space).

Example

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

Output: One possible answer could be [1, 6, 1, 5, 1, 4].

Clarifying Questions

1. Q: Can the input be empty? A: No. You can assume that the array will have at least one element.

2. Q: What should be the behavior in case of ties (duplicate elements)? A: The input array can have duplicate numbers, and the resultant order should still satisfy the nums[0] < nums[1] > nums[2] < nums[3]... condition.

3. Q: Can additional space be used for processing? A: The solution should be done in-place. However, constant extra space is allowed.

Strategy

1. Sort the Array: Start by sorting the array.
2. Split the Array: Split the sorted array into two halves.
3. Interleave Elements: Interleave elements from the two halves such that the elements from the first half are placed in even indices, and elements from the second half are placed in odd indices.

Here’s a more detailed step-by-step approach:

1. Sort the Array: Sorting the array will help in easily accessing the smallest and largest available elements.
2. Split into Two Halves: Split the sorted array into two halves, the smaller half and the larger half. Ensure the larger half is not smaller than the smaller half.
3. Reverse and Reallocate: Reverse the smaller half and larger half, then reassign elements to form the wiggle sequence.

Code

def wiggleSort(nums):
    nums.sort()
    half = len(nums[::2])
    nums[::2], nums[1::2] = nums[:half][::-1], nums[half:][::-1]

# Example usage:
nums = [1, 5, 1, 1, 6, 4]
wiggleSort(nums)
print(nums)  # Output might be [1, 6, 1, 5, 1, 4] or [1, 6, 1, 5, 1, 4]

Time Complexity

• Time Complexity: O(n log n) for sorting the array.
• Space Complexity: O(1) since we are doing the sorting and interleaving in-place without using additional arrays or lists. The slicing operation in Python essentially references the array and does not count as extra space.

Note: The slicing operation appears to use extra space, but since it’s being done in a manner that rearranges elements in-place, it is constant in terms of additional space usage, adhering to in-place requirements.

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