In this guide, you will get 376. Wiggle Subsequence LeetCode Solution with the best time and space complexity. The solution to Wiggle Subsequence problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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).
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
Follow up: Could you solve this in O(n) time?
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
376. Wiggle Subsequence LeetCode Solution in C++
class Solution {
public:
int wiggleMaxLength(vector<int>& nums) {
int increasing = 1;
int decreasing = 1;
for (int i = 1; i < nums.size(); ++i)
if (nums[i] > nums[i - 1])
increasing = decreasing + 1;
else if (nums[i] < nums[i - 1])
decreasing = increasing + 1;
return max(increasing, decreasing);
}
}; /* code provided by PROGIEZ */
376. Wiggle Subsequence LeetCode Solution in Java
class Solution {
public int wiggleMaxLength(int[] nums) {
int increasing = 1;
int decreasing = 1;
for (int i = 1; i < nums.length; ++i)
if (nums[i] > nums[i - 1])
increasing = decreasing + 1;
else if (nums[i] < nums[i - 1])
decreasing = increasing + 1;
return Math.max(increasing, decreasing);
}
} // code provided by PROGIEZ
376. Wiggle Subsequence LeetCode Solution in Python
class Solution:
def wiggleMaxLength(self, nums: list[int]) -> int:
increasing = 1
decreasing = 1
for a, b in itertools.pairwise(nums):
if b > a:
increasing = decreasing + 1
elif b < a:
decreasing = increasing + 1
return max(increasing, decreasing) # code by PROGIEZ
