376. Wiggle Subsequence

A sequence of numbers is called a wiggle sequence if 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 fewer than two elements is trivially a wiggle sequence.

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

Given a sequence of integers, return the length of the longest subsequence that is a wiggle sequence. A subsequence is obtained by deleting some number of elements (eventually, also zero) from the original sequence, leaving the remaining elements in their original order.

Example 1:

Input: [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence.

Example 2:

Input: [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].

Example 3:

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

Solution

Approach 1: DP.

Approach 2: greedy.

Code (Python)

Approach 1:

    def wiggleMaxLength(self, nums: List[int]) -> int:
        # DP -- keep 2 tables, for the current index being the peak and being the valley
        if not nums:
            return 0
        if len(nums) == 1:
            return 1
        as_peak, as_valley = [1] * len(nums), [1] * len(nums)
        global_max = 2
        for i in range(1, len(nums)):
            for j in range(i):
                if nums[j] < nums[i]:
                    as_peak[i] = max(as_peak[i], as_valley[j] + 1)
                    # don't have to update as_valley because it's never going to exceed the previous ones
                elif nums[j] > nums[i]:
                    as_valley[i] = max(as_valley[i], as_peak[j] + 1)
                # again don't have to update as_peak or as_valley for ==
        return max(max(as_peak), max(as_valley))

Approach 2:

    def wiggleMaxLength(self, nums: List[int]) -> int:
        # greedy: think of nums as a stack, we need to go through the nums and get the max and min peak, and count the alternations
        if len(nums) <= 1:
            return len(nums)
        if len(nums) == 2:
            return 1 if nums[0] == nums[1] else 2
        diffs = list(filter(lambda x: x != 0, [nums[i] - nums[i-1] for i in range(1, len(nums))]))
        if not diffs:
            return 1
        max_len = 1
        for i in range(1, len(diffs)):
            if diffs[i] * diffs[i-1] < 0:
                max_len += 1
        return max_len + 1

Code (C++)

Approach 1:

Approach 2: