Last updated on February 9th, 2025 at 07:26 am

Here, we see a Wiggle Sort II LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

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Wiggle Sort II LeetCode Solution

1. Problem Statement

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

2. Coding Pattern Used in Solution

The coding pattern used in all the provided implementations is “Sorting and Index Manipulation”, a common approach in problems where the goal is to rearrange elements in a specific order based on their sorted positions. The idea is to sort the array and then rearrange elements into a “wiggle” pattern, where elements alternate between smaller and larger values.

3. Code Implementation in Different Languages

3.1 Wiggle Sort II C++

class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        vector<int> sorted(nums);
        sort(sorted.begin(), sorted.end());
        for (int i=nums.size()-1, j=0, k=i/2+1; i>=0; i--)
            nums[i] = sorted[i&1 ? k++ : j++];
    }
};

3.2 Wiggle Sort II Java

class Solution {
    public void wiggleSort(int[] nums) {
       int n=nums.length-1;
       int[] newarr=Arrays.copyOf(nums,nums.length);
       Arrays.sort(newarr);
        for(int i=1;i<nums.length;i+=2)
            nums[i]=newarr[n--];
        for(int i=0;i<nums.length;i+=2)
            nums[i]=newarr[n--];
    }
}

3.3 Wiggle Sort II JavaScript

var wiggleSort = function(nums) {
    nums.sort((b,a)=>b-a);
    let mid = Math.floor(nums.length/2)
    mid+=nums.length%2==0?0:1;
    let even = nums.slice(0, mid);
    let odd = nums.slice(mid);
    for(let i=0;i<nums.length;i++) {
        if (i%2==0) {
            nums[i] = even.pop();
        } else {
            nums[i] = odd.pop();
        }
    }
};

3.4 Wiggle Sort II Python

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

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(n log n)	O(n)
Java	O(n log n)	O(n)
JavaScript	O(n log n)	O(n)
Python	O(n log n)	O(n)