#324

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Wiggle Sort II

Rearrange an array in a way that every odd-indexed element is greater than its adjacent even-indexed elements.

array divide and conquer greedy sorting quickselect

Problem Statement

Given an integer array nums, reorder it in such a way that nums[0] <= nums[1] >= nums[2] <= nums[3]... For example, if the input is [1,5,1,1,6,4], the reordered array should be [1,6,1,5,1,4] (although [1,4,1,5,1,6] is also acceptable).

You are guaranteed that a valid answer exists for the given input. The task emphasizes finding an efficient solution where each element's position respects the 'wiggle' condition. Make sure to consider a greedy approach and invariant validation for the solution to be optimal.

Examples

Example 1

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

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

[1,4,1,5,1,6] is also accepted.

Example 2

Input: nums = [1,3,2,2,3,1]

Output: [2,3,1,3,1,2]

Example details omitted.

Constraints

1 <= nums.length <= 5 * 104
0 <= nums[i] <= 5000
It is guaranteed that there will be an answer for the given input nums.

Solution Approach

Greedy Choice and Invariant Validation

The solution starts by focusing on the greedy choice to place the largest elements in odd indices, ensuring that every odd-indexed element is greater than its adjacent even-indexed elements. Then, after sorting the array, the elements are placed carefully to satisfy the 'wiggle' condition.

Array Sorting and Element Placement

By first sorting the array, we can easily select elements to position correctly at odd and even indices. Sorting helps simplify the arrangement process by ensuring that the largest elements are placed at the odd indices first, fulfilling the 'wiggle' condition with minimal swaps.

Efficient Invariant Check

After the array is rearranged, an invariant check is performed to ensure that each element at an odd index is greater than its neighbors. This check validates whether the 'wiggle' condition holds true across the entire array.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution depends on the approach used to sort the array. If using Quickselect or a similar sorting technique, the time complexity is O(n log n), while the space complexity varies based on the specific approach but generally remains O(n) for storing temporary results or auxiliary data.

What Interviewers Usually Probe

Assess how the candidate applies the greedy approach to solve the problem efficiently.
Evaluate the candidate's ability to manage array manipulations and element positioning for this pattern.
Check for understanding of the 'wiggle' condition and whether the candidate can optimize the process using sorting or selective placement.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the solution by trying to compare every adjacent pair rather than focusing on the greedy choice and invariant validation.
Failing to correctly handle the array sorting step, which is essential for ensuring the correct wiggle order.
Not performing a final check after rearranging the array to ensure that the wiggle condition holds for all elements.

Follow-up variants

Apply the same approach to larger datasets or arrays with different constraints for performance evaluation.
Implement a solution with an alternative sorting method or use partition-based techniques like Quickselect.
Test the approach by adding additional constraints, such as ensuring specific values in the array are placed at certain indices.

FAQ

What is the key pattern in the Wiggle Sort II problem?

The key pattern is greedy choice plus invariant validation, where you place the largest elements in odd indices and validate the arrangement.

How does sorting help in solving this problem?

Sorting the array first ensures that the largest elements can be placed in odd indices, helping to satisfy the wiggle condition with minimal manipulation.

Is the wiggle condition required for all elements in the array?

Yes, the wiggle condition must hold true for every adjacent pair of elements, meaning each odd-indexed element must be greater than its neighbors.

How do I avoid common mistakes in this problem?

Ensure that the array is sorted correctly and that you check the final arrangement to verify that the wiggle condition is satisfied for every element.

What are some variations of the Wiggle Sort II problem?

Variations include applying the approach to arrays with additional constraints or exploring performance with larger datasets using alternative sorting methods.

terminal

Solution

Solution 1

#### Python3

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        arr = sorted(nums)
        n = len(arr)
        i, j = (n - 1) >> 1, n - 1
        for k in range(n):
            if k % 2 == 0:
                nums[k] = arr[i]
                i -= 1
            else:
                nums[k] = arr[j]
                j -= 1

Solution 2

#### Python3

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        arr = sorted(nums)
        n = len(arr)
        i, j = (n - 1) >> 1, n - 1
        for k in range(n):
            if k % 2 == 0:
                nums[k] = arr[i]
                i -= 1
            else:
                nums[k] = arr[j]
                j -= 1

Continue Practicing

#347 top k frequent elements #435 non overlapping intervals #455 assign cookies