How do you solve Wiggle Subsequence on LeetCode?

Wiggle Subsequence (LeetCode #376) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Wiggle sequences can be identified by the sign of differences between consecutive elements.

What is the brute force solution for Wiggle Subsequence?

The brute force approach for Wiggle Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the array and checking each one to see if it forms a wiggle sequence. This is simple but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Wiggle Subsequence?

Wiggle Subsequence uses the following concepts: Array, Dynamic Programming, Greedy. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Wiggle Subsequence?

Always consider edge cases first; they can help you avoid simple mistakes. Explain your thought process clearly while coding; it shows your understanding. If you get stuck, try to simplify the problem or break it down into smaller parts.

What are common mistakes when solving Wiggle Subsequence?

Failing to handle edge cases like arrays of length 1 or 2. Not correctly alternating between counting up and down in the optimal solution.

#376

Wiggle Subsequence

Medium

ArrayDynamic ProgrammingGreedyDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses dynamic programming to keep track of the lengths of wiggle subsequences ending at each index. This is efficient because we only need to look at the previous element to determine if we can extend the wiggle sequence.

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Algorithm

3 steps

1Step 1: Initialize two variables, 'up' and 'down', to track the length of wiggle subsequences ending with an upward or downward difference.
2Step 2: Iterate through the array, comparing each element with the previous one to determine if the difference is positive or negative.
3Step 3: Update 'up' or 'down' based on the comparison, ensuring to alternate the counts.

solution.py15 lines

1# Full working Python code
2
3def wiggle_length(nums):
4    if len(nums) < 2:
5        return len(nums)
6    up = down = 1
7    for i in range(1, len(nums)):
8        if nums[i] > nums[i - 1]:
9            up = down + 1
10        elif nums[i] < nums[i - 1]:
11            down = up + 1
12    return max(up, down)
13
14# Example usage
15print(wiggle_length([1, 7, 4, 9, 2, 5]))

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array, and the space complexity is O(1) since we use only a fixed amount of extra space.

1Wiggle sequences can be identified by the sign of differences between consecutive elements.
2Dynamic programming can efficiently track the length of sequences without generating all subsequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.