18. Wiggle Subsequence.cpp

/*
Wiggle Subsequence
==================

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

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

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

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].

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?
*/

class Solution
{
public:
  int wiggleMaxLength(vector<int> &nums)
  {
    if (nums.size() == 2)
    {
      if (nums[0] == nums[1])
        return 1;
      else
        return 2;
    }

    int ans = 1;
    int n = nums.size(), i = 1;

    while (i < n && nums[i] == nums[i - 1])
      i++;
    if (i == n)
      return ans;

    bool prev_diff_pos = nums[i] - nums[0] > 0 ? true : false;
    int last = nums[i];
    ans++;
    i++;

    while (i < n)
    {
      if ((prev_diff_pos && nums[i] < last) || (!prev_diff_pos && nums[i] > last))
      {
        ans++;
        prev_diff_pos = !prev_diff_pos;
      }
      last = nums[i];
      i++;
    }

    return ans;
  }
};