Wiggle Subsequence – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

A wiggle sequence is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.

• For example, [1, 7, 4, 9, 2, 5] is a wiggle sequence because the differences (6, -3, 5, -7, 3) alternate between positive and negative.
• In contrast, [1, 4, 7, 2, 5] and [1, 7, 4, 5, 5] are not wiggle sequences. The first is not because its first two differences are positive, and the second is not because its last difference is zero.

A subsequence is obtained by deleting some elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.

Given an integer array nums, return the length of the longest wiggle subsequence of nums.

Example 1:

Input: nums = [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence with differences (6, -3, 5, -7, 3).

Example 2:

Input: nums = [1,17,5,10,13,15,10,5,16,8]
Output: 7
Explanation: There are several subsequences that achieve this length.
One is [1, 17, 10, 13, 10, 16, 8] with differences (16, -7, 3, -3, 6, -8).

Example 3:

Input: nums = [1,2,3,4,5,6,7,8,9]
Output: 2

Constraints:

• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 1000

C++ Wiggle Subsequence LeetCode Solution

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int size=nums.size(), peak=1, valley=1;
        for(int i=1; i<size; ++i){
                 if(nums[i]>nums[i-1]) peak = valley + 1;
            else if(nums[i]<nums[i-1]) valley = peak + 1;
        }
        return max(peak , valley );
    }
};

Java Wiggle Subsequence LeetCode Solution

 public class Solution {
	public int wiggleMaxLength(int[] nums) {
		if (nums.length == 0 || nums.length == 1) {
			return nums.length;
		}
		int k = 0;
		while (k < nums.length - 1 && nums[k] == nums[k + 1]) {  //Skips all the same numbers from series beginning eg 5, 5, 5, 1
			k++;
		}
		if (k == nums.length - 1) {
			return 1;
		}
		int result = 2;     // This will track the result of result array
		boolean smallReq = nums[k] < nums[k + 1];       //To check series starting pattern
		for (int i = k + 1; i < nums.length - 1; i++) {
			if (smallReq && nums[i + 1] < nums[i]) {
				nums[result] = nums[i + 1];
				result++;
				smallReq = !smallReq;    //Toggle the requirement from small to big number
			} else {
				if (!smallReq && nums[i + 1] > nums[i]) {
					nums[result] = nums[i + 1];
					result++;
					smallReq = !smallReq;    //Toggle the requirement from big to small number
				}
			}
		}
		return result;
	}
}
 

Python 3 Wiggle Subsequence LeetCode Solution

class Solution:
    def wiggleMaxLength(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        
        length = 1
        up = None # current is increasing or not
        for i in range(1, len(nums)):
            if nums[i] > nums[i - 1] and up != True:
                length += 1
                up = True
            if nums[i] < nums[i - 1] and up != False:
                length += 1
                up = False
        return length

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