376. Wiggle Subsequence¶

Difficulty: Medium

376. Wiggle Subsequence

Medium

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

Follow up: Could you solve this in O(n) time?

Solution¶

class Solution {
    private int dp[][][];
    public int wiggleMaxLength(int[] nums) {
        int n = nums.length;
        dp = new int[n + 1][n + 1][3];
        for (int current[][] : dp) for (int current1[] : current) Arrays.fill(current1, -1);
        int res = solve(0, -1, -1, nums);
        return res;
    }

    private int solve(int ind, int prev_diff, int prev_ind, int arr[]) {
        if (ind >= arr.length) return 0;
        if (dp[ind][prev_ind + 1][prev_diff + 1] != -1) return dp[ind][prev_ind + 1][prev_diff + 1];
        if (prev_ind == -1) {
            int op1 = 0, op2 = 0;
            op1 = 1 + solve(ind + 1, prev_diff, ind, arr);
            op2 = solve(ind + 1, prev_diff, prev_ind, arr);
            return dp[ind][prev_ind + 1][prev_diff + 1] = Math.max(op1, op2);
        } 
        else if (prev_diff == -1) {
            int op1 = 0, op2 = 0;
            if (arr[ind] - arr[prev_ind] > 0) op1 = 1 + solve(ind + 1, 1, ind, arr);
            if (arr[ind] - arr[prev_ind] < 0) op1 = 1 + solve(ind + 1, 0, ind, arr); 
            op2 = solve(ind + 1, prev_diff, prev_ind, arr);
            return dp[ind][prev_ind + 1][prev_diff + 1] = Math.max(op1, op2);
        }
        int op1 = 0, op2 = 0;
        if (arr[ind] - arr[prev_ind] > 0) {
            if (prev_diff == 0) op1 = 1 + solve(ind + 1, 1, ind, arr);
            else return 0;
        }
        if (arr[ind] - arr[prev_ind] < 0) {
            if (prev_diff == 1) op1 = 1 + solve(ind + 1, 0, ind, arr);
            else return 0;
        }
        op2 = solve(ind + 1, prev_diff, prev_ind, arr);
        return dp[ind][prev_ind + 1][prev_diff + 1] = Math.max(op1, op2);
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here