Wiggle Sort

280.Wiggle Sort

Array

Greedy

Bit Manipulation

Problem Statement:

Given an unsorted array nums, reorder it in-place such that nums[0] ≤ nums[1] ≥ nums[2] ≤ nums[3].... The final array should satisfy the wiggle property where every odd index is greater than or equal to its adjacent elements.

Example:

Input:

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

→

Output:

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

Algorithm:

1. Traverse array checking adjacent elements
2. Even indices should be less than next
3. Odd indices should be greater than next
4. Swap if condition not met

Complexity:

Time: O(n) | Space: O(1)

Java Solution:

public void wiggleSort(int[] nums) {
    for (int i = 0; i < nums.length - 1; i++) {
        if (((i % 2 == 0) && nums[i] > nums[i + 1])  // Even index should be less than next
                || ((i % 2 == 1) && nums[i] < nums[i + 1])) {  // Odd index should be greater than next
            swap(nums, i, i + 1);
        }
    }
}

public void swap(int[] nums, int i, int j) {
    nums[i] ^= nums[j];
    nums[j] ^= nums[i];
    nums[i] ^= nums[j];
}

Python Solution:

def wiggleSort(self, nums: List[int]) -> None:
    for i in range(len(nums) - 1):
        if ((i % 2 == 0 and nums[i] > nums[i + 1]) or
            (i % 2 == 1 and nums[i] < nums[i + 1])):
            nums[i], nums[i + 1] = nums[i + 1], nums[i]  # Python swap

C++ Solution:

class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        for (int i = 0; i < nums.size() - 1; i++) {
            if ((i % 2 == 0 && nums[i] > nums[i + 1]) ||
                (i % 2 == 1 && nums[i] < nums[i + 1])) {
                swap(nums[i], nums[i + 1]);  // C++ built-in swap
            }
        }
    }
};