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Formatted question description: https://leetcode.ca/all/376.html

 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


 Follow up:
 Can you do it in O(n) time?

 @tag-dp

Algorithm

DP

We maintain two dp arrays p and q,

  • p[i] indicates the maximum length of the swing subsequence with a positive first difference when the position i is reached
  • q[i] indicates the maximum length of the wobble sub-sequence with a negative first difference when reaching the i position

We traverse the array from i=1, and then for each traversed number, traverse to this number from the beginning position, then compare nums[i] and nums[j], and update the corresponding positions respectively.

Greedy

Maintain two variables p and q, and then traverse the array

  • If the current number is greater than the previous number, then p=q+1
  • If the current number is smaller than the previous number, then q=p+1

Finally, compare the larger value of p and q with n, and choose the smaller one.

Code

Java

  • import java.util.Arrays;

public class Wiggle_Subsequence {

    public static void main(String[] args) {
        Wiggle_Subsequence out = new Wiggle_Subsequence();
        Solution s = out.new Solution();

        System.out.println(s.wiggleMaxLength(new int[]{1,17,5,10,13,15,10,5,16,8}));
    }

    public class Solution_greedy {
        public int wiggleMaxLength(int[] nums) {
            if (nums.length < 2) {
                return nums.length;
            }

            int up = 1, down = 1, n = nums.length;
            for (int i = 1; i < n; ++i) { // start from 2nd
                if (nums[i] > nums[i - 1]) up = down + 1;
                else if (nums[i] < nums[i - 1]) down = up + 1;
            }
            return Math.min(n, Math.max(up, down));
        }
    }

    public class Solution_better_oN {
        public int wiggleMaxLength(int[] nums) {
            if (nums.length < 2) {
                return nums.length;
            }

            int n = nums.length;
            int[] up = new int[n];
            int[] down = new int[n];
            up[0] = down[0] = 1;
            for (int i = 1; i < n; i++) {
                if (nums[i] > nums[i - 1]) {
                    up[i] = down[i - 1] + 1;
                    down[i] = down[i - 1];
                } else if (nums[i] < nums[i - 1]) {
                    down[i] = up[i - 1] + 1;
                    up[i] = up[i - 1];
                } else { // @note: test case, input: [0,0]
                    down[i] = down[i - 1];
                    up[i] = up[i - 1];
                }
            }
            return Math.max(down[n - 1], up[n - 1]);
        }
    }

    public class Solution { // o(N^2)

        // @note: could start with either up or down, both valid
        public int wiggleMaxLength(int[] nums) {

            if (nums.length < 2) {
                return nums.length;
            }

            // up[i] refers to the length of the longest wiggle subsequence obtained so far
            //      considering i^th element as the last element of the wiggle subsequence and ending with a rising wiggle.
            int[] up = new int[nums.length];

            int[] down = new int[nums.length];

            Arrays.fill(up, 1); // initialization, each position is at least 1 for that char
            Arrays.fill(down, 1); // @note: or don't do this step, in final return just `1+()`

            for (int i = 1; i < nums.length; i++) {
                for(int j = 0; j < i; j++) {
                    if (nums[i] > nums[j]) {
                        up[i] = Math.max(up[i],down[j] + 1);
                    } else if (nums[i] < nums[j]) {
                        down[i] = Math.max(down[i],up[j] + 1);
                    }
                }
            }

            return Math.max(down[nums.length - 1], up[nums.length - 1]);
        }
    }
}

  • // OJ: https://leetcode.com/problems/wiggle-subsequence/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int inc = 1, dec = 1, N = nums.size();
        for (int i = 1; i < N; ++i) {
            if (nums[i] > nums[i - 1]) inc = dec + 1;
            else if (nums[i] < nums[i - 1]) dec = inc + 1;
        }
        return max(inc, dec);
    }
};

  • class Solution:
    def wiggleMaxLength(self, nums: List[int]) -> int:
        up = down = 1
        for i in range(1, len(nums)):
            if nums[i] > nums[i - 1]:
                up = max(up, down + 1)
            elif nums[i] < nums[i - 1]:
                down = max(down, up + 1)
        return max(up, down)

############

class Solution(object):
  def wiggleMaxLength(self, nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    if not nums:
      return 0
    up = down = 1
    for i in range(1, len(nums)):
      if nums[i] > nums[i - 1]:
        up = down + 1
      elif nums[i] < nums[i - 1]:
        down = up + 1
    return max(up, down)

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