0324. Wiggle Sort II

https://leetcode.com/problems/wiggle-sort-ii

Description

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?

ac1: quick select

1. Find kth element, same with https://leetcode.com/problems/kth-largest-element-in-an-array/description/

2. Difficult part is how to rearrange the new array. See explanation.

class Solution {
    public void wiggleSort(int[] nums) {
        // edge case
        if (nums == null || nums.length == 0) return;

        int n = nums.length;
        int mid = kthElement(nums, (n+1)/2); // find the middle point, O(n) time
        int[] copy = new int[n];

        // bigger -> odd, left to right, smaller -> even, right to left
        int odd = 1, even = n % 2 == 0 ? n - 2 : n - 1;
        for (int i = 0; i < n; i++) {
            if (nums[i] > mid) {
                copy[odd] = nums[i];
                odd += 2;
            }
            if (nums[i] < mid) {
                copy[even] = nums[i];
                even -= 2;
            }
        }

        // fill mid to the rest. Deal with equal elements.
        while (odd < n) {
            copy[odd] = mid;
            odd += 2;
        }
        while (even >= 0) {
            copy[even] = mid;
            even -= 2;
        }

        // copy back
        for (int i = 0; i < n; i++) {
            nums[i] = copy[i];
        }

    }

    private int kthElement(int[] nums, int k) {
        int l = 0, r = nums.length - 1, res = 0;
        while (l <= r) {
            int cut = partition(nums, l, r);
            if (cut + 1 == k) {
                res = nums[cut];
                break;
            } else if (k < cut + 1) {
                r = cut - 1;
            } else {
                l = cut + 1;
            }
        }
        return res;
    }
    private int partition(int[] nums, int start, int end) {
        if (start == end) return start;
        int pivot = nums[start], l = start + 1, r = end;
        while (l <= r) {
            while (l <= r && nums[l] > pivot) l++;
            while (l <= r && nums[r] < pivot) r--;
            if (l <= r) {
                swap(nums, l, r);
                l++;
                r--;
            }
        }
        swap(nums, start, r);
        return r;
    }
    private void swap(int[] nums, int a, int b) {
        int tmp = nums[a];
        nums[a] = nums[b];
        nums[b] = tmp;
    }
}

/*
O(N) time and space
*/

ac2: sort

O(nlogn) time, but it's faster than previous solution, weird

class Solution {
    public void wiggleSort(int[] nums) {
        // edge case
        if (nums == null || nums.length == 0) return;

        int n = nums.length;
        Arrays.sort(nums);
        int[] copy = nums.clone();


        int i = n - 1;
        int idx = 1;
        while (idx < n) {
            nums[idx] = copy[i];
            idx += 2;
            i--;
        }
        idx = 0;
        while (idx < n) {
            nums[idx] = copy[i];
            idx += 2;
            i--;
        }

    }

}
// O(NlogN) time, O(N) space

ac3: index mapping

0(N) time, O(1) space. But it's insanely hard to understand. See this post: https://leetcode.com/problems/wiggle-sort-ii/discuss/77682/step-by-step-explanation-of-index-mapping-in-java.