Wiggle Sort II LeetCode Solution – Queslers

Problem – Wiggle Sort II

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]

Constraints:

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

Follow Up: Can you do it in O(n) time and/or in-place with O(1) extra space?

Wiggle Sort II LeetCode Solution in Java

   public void wiggleSort(int[] nums) {
        int median = findKthLargest(nums, (nums.length + 1) / 2);
        int n = nums.length;

        int left = 0, i = 0, right = n - 1;

        while (i <= right) {

            if (nums[newIndex(i,n)] > median) {
                swap(nums, newIndex(left++,n), newIndex(i++,n));
            }
            else if (nums[newIndex(i,n)] < median) {
                swap(nums, newIndex(right--,n), newIndex(i,n));
            }
            else {
                i++;
            }
        }


    }

    private int newIndex(int index, int n) {
        return (1 + 2*index) % (n | 1);
    }

Wiggle Sort II LeetCode Solution in Python

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

Wiggle Sort II LeetCode Solution in C++

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++];
}

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