280. Wiggle Sort


Given an unsorted array nums, reorder it in-place such that nums[0] <= nums[1] >= nums[2] <= nums[3]....

For example, given nums = [3, 5, 2, 1, 6, 4], one possible answer is [1, 6, 2, 5, 3, 4].


Solution

Approach #1 (Sorting) [Accepted]

The obvious solution is to just sort the array first, then swap elements pair-wise starting from the second element. For example:

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

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

public void wiggleSort(int[] nums) {
    Arrays.sort(nums);
    for (int i = 1; i < nums.length - 1; i += 2) {
        swap(nums, i, i + 1);
    }
}

private void swap(int[] nums, int i, int j) {
    int temp = nums[i];
    nums[i] = nums[j];
    nums[j] = temp;
}

Complexity analysis

  • Time complexity : . The entire algorithm is dominated by the sorting step, which costs time to sort elements.

  • Space complexity : . Space depends on the sorting implementation which, usually, costs auxiliary space if heapsort is used.

Approach #2 (One-pass Swap) [Accepted]

Intuitively, we should be able to reorder it in one-pass. As we iterate through the array, we compare the current element to its next element and if the order is incorrect, we swap them.

public void wiggleSort(int[] nums) {
    boolean less = true;
    for (int i = 0; i < nums.length - 1; i++) {
        if (less) {
            if (nums[i] > nums[i + 1]) {
                swap(nums, i, i + 1);
            }
        } else {
            if (nums[i] < nums[i + 1]) {
                swap(nums, i, i + 1);
            }
        }
        less = !less;
    }
}

We could shorten the code further by compacting the condition to a single line. Also observe the boolean value of less actually depends on whether the index is even or odd.

public void wiggleSort(int[] nums) {
    for (int i = 0; i < nums.length - 1; i++) {
        if (((i % 2 == 0) && nums[i] > nums[i + 1])
                || ((i % 2 == 1) && nums[i] < nums[i + 1])) {
            swap(nums, i, i + 1);
        }
    }
}

Here is another amazing solution by @StefanPochmann who came up with originally here.

public void wiggleSort(int[] nums) {
    for (int i = 0; i < nums.length - 1; i++) {
        if ((i % 2 == 0) == (nums[i] > nums[i + 1])) {
            swap(nums, i, i + 1);
        }
    }
}

Complexity analysis

  • Time complexity : . In the worst case we swap at most times. An example input is [2,1,3,1,4,1].

  • Space complexity : .

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