How do you solve Wiggle Sort II on LeetCode?

Wiggle Sort II (LeetCode #324) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting helps in easily accessing the smallest and largest elements.

What is the time complexity of Wiggle Sort II?

The optimal solution for Wiggle Sort II runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for the result array.

What is the brute force solution for Wiggle Sort II?

The brute force approach for Wiggle Sort II is Brute Force, with O(n²) time complexity and O(n!) space complexity. The brute-force approach involves generating all possible permutations of the array and checking each one to see if it satisfies the wiggle sort condition. This is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Wiggle Sort II?

Wiggle Sort II uses the following concepts: Array, Divide and Conquer, Greedy, Sorting, Quickselect. The recommended patterns to study are: Sorting, Two Pointers, Array.

What are the interview tips for Wiggle Sort II?

Explain your thought process clearly while coding. Discuss the time and space complexity of your approach. Be prepared to optimize your solution if asked.

What are common mistakes when solving Wiggle Sort II?

Not considering edge cases where the array has duplicate elements. Forgetting to check the wiggle condition properly.

#324

Wiggle Sort II

Medium

ArrayDivide and ConquerGreedySortingQuickselectSortingTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n!)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution involves sorting the array and then cleverly rearranging it using a two-pointer technique. This ensures that we can achieve the desired wiggle sort in linear time while maintaining the required conditions.

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Algorithm

4 steps

1Step 1: Sort the array.
2Step 2: Create a new array of the same length to hold the wiggle sorted result.
3Step 3: Use two pointers to fill the new array: one for the larger half and one for the smaller half.
4Step 4: Place elements in the new array according to the wiggle pattern.

solution.py17 lines

1# Full working Python code
2def wiggleSort(nums):
3    nums.sort()
4    n = len(nums)
5    result = [0] * n
6    left, right = (n + 1) // 2 - 1, n - 1
7    for i in range(n):
8        if i % 2 == 0:
9            result[i] = nums[left]
10            left -= 1
11        else:
12            result[i] = nums[right]
13            right -= 1
14    return result
15
16# Example usage
17print(wiggleSort([1, 5, 1, 1, 6, 4]))

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for the result array.

1Sorting helps in easily accessing the smallest and largest elements.
2Using two pointers allows us to fill the result array in a structured manner.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.