LeetCode #376 — MEDIUM

Wiggle Subsequence

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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?

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Greedy

Example 1

[1,7,4,9,2,5]

Example 2

[1,17,5,10,13,15,10,5,16,8]

Example 3

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

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #376: Wiggle Subsequence
class Solution {
    public int wiggleMaxLength(int[] nums) {
        int n = nums.length;
        int ans = 1;
        int[] f = new int[n];
        int[] g = new int[n];
        f[0] = 1;
        g[0] = 1;
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[j] < nums[i]) {
                    f[i] = Math.max(f[i], g[j] + 1);
                } else if (nums[j] > nums[i]) {
                    g[i] = Math.max(g[i], f[j] + 1);
                }
            }
            ans = Math.max(ans, Math.max(f[i], g[i]));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #376: Wiggle Subsequence
func wiggleMaxLength(nums []int) int {
	n := len(nums)
	f := make([]int, n)
	g := make([]int, n)
	f[0], g[0] = 1, 1
	ans := 1
	for i := 1; i < n; i++ {
		for j := 0; j < i; j++ {
			if nums[j] < nums[i] {
				f[i] = max(f[i], g[j]+1)
			} else if nums[j] > nums[i] {
				g[i] = max(g[i], f[j]+1)
			}
		}
		ans = max(ans, max(f[i], g[i]))
	}
	return ans
}

# Accepted solution for LeetCode #376: Wiggle Subsequence
class Solution:
    def wiggleMaxLength(self, nums: List[int]) -> int:
        n = len(nums)
        ans = 1
        f = [1] * n
        g = [1] * n
        for i in range(1, n):
            for j in range(i):
                if nums[j] < nums[i]:
                    f[i] = max(f[i], g[j] + 1)
                elif nums[j] > nums[i]:
                    g[i] = max(g[i], f[j] + 1)
            ans = max(ans, f[i], g[i])
        return ans

// Accepted solution for LeetCode #376: Wiggle Subsequence
struct Solution;
use std::cmp::Ordering::*;

impl Solution {
    fn wiggle_max_length(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        if n == 0 {
            return 0;
        }
        let mut up: Vec<usize> = vec![1];
        let mut down: Vec<usize> = vec![1];
        for i in 1..n {
            match nums[i].cmp(&nums[i - 1]) {
                Greater => {
                    down.push(up[i - 1] + 1);
                    up.push(up[i - 1]);
                }
                Less => {
                    down.push(down[i - 1]);
                    up.push(down[i - 1] + 1);
                }
                Equal => {
                    down.push(down[i - 1]);
                    up.push(up[i - 1]);
                }
            }
        }
        up.into_iter().chain(down.into_iter()).max().unwrap() as i32
    }
}

#[test]
fn test() {
    let nums = vec![1, 7, 4, 9, 2, 5];
    let res = 6;
    assert_eq!(Solution::wiggle_max_length(nums), res);
    let nums = vec![1, 17, 5, 10, 13, 15, 10, 5, 16, 8];
    let res = 7;
    assert_eq!(Solution::wiggle_max_length(nums), res);
    let nums = vec![1, 2, 3, 4, 5, 6, 7, 8, 9];
    let res = 2;
    assert_eq!(Solution::wiggle_max_length(nums), res);
}

// Accepted solution for LeetCode #376: Wiggle Subsequence
function wiggleMaxLength(nums: number[]): number {
    const n = nums.length;
    const f: number[] = Array(n).fill(1);
    const g: number[] = Array(n).fill(1);
    let ans = 1;
    for (let i = 1; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            if (nums[i] > nums[j]) {
                f[i] = Math.max(f[i], g[j] + 1);
            } else if (nums[i] < nums[j]) {
                g[i] = Math.max(g[i], f[j] + 1);
            }
        }
        ans = Math.max(ans, f[i], g[i]);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.