Solution Explanation for Wiggle Sort

The problem asks to reorder an integer array nums such that nums[0] <= nums[1] >= nums[2] <= nums[3].... This means we need an alternating pattern of less than or equal to and greater than or equal to.

Approach

The most straightforward approach is a greedy one. We iterate through the array, comparing each element to its preceding element. If the pattern is violated (e.g., nums[i] < nums[i-1] when i is odd, or nums[i] > nums[i-1] when i is even), we swap the elements to restore the pattern. This ensures that after the loop, the array satisfies the wiggle sort condition.

Code Explanation (Python)

class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        for i in range(1, len(nums)):
            if (i % 2 == 1 and nums[i] < nums[i - 1]) or (
                i % 2 == 0 and nums[i] > nums[i - 1]
            ):
                nums[i], nums[i - 1] = nums[i - 1], nums[i]

The Python code directly implements the greedy approach. The for loop iterates from the second element. The if condition checks if the current element violates the wiggle sort pattern based on its index (i % 2). If a violation is found, a simple swap using tuple assignment (nums[i], nums[i - 1] = nums[i - 1], nums[i]) corrects it. The function modifies the input array in-place, as specified.

Code Explanation (Java)

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

The Java code is very similar to the Python version. It uses a helper function swap to improve readability. The logic for checking the pattern and swapping elements remains the same.

Code Explanation (C++)

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

The C++ code uses std::swap for efficient swapping, but the overall algorithm remains consistent with the Python and Java versions.

Code Explanation (Go)

func wiggleSort(nums []int) {
	for i := 1; i < len(nums); i++ {
		if (i%2 == 1 && nums[i] < nums[i-1]) || (i%2 == 0 && nums[i] > nums[i-1]) {
			nums[i], nums[i-1] = nums[i-1], nums[i]
		}
	}
}

The Go code is a concise implementation, similar in structure to the Python code. It directly uses the tuple assignment for swapping.

Time and Space Complexity

• Time Complexity: O(n), where n is the length of the input array. We iterate through the array once.
• Space Complexity: O(1). The algorithm operates in-place; it doesn't use any extra space that grows with the input size. Only a constant amount of extra space is used for variables.