How do you solve Arithmetic Slices II - Subsequence on LeetCode?

Arithmetic Slices II - Subsequence (LeetCode #446) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using hashmaps allows for efficient counting of subsequences.

What is the brute force solution for Arithmetic Slices II - Subsequence?

The brute force approach for Arithmetic Slices II - 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 is an arithmetic sequence. This is straightforward but inefficient, as it can lead to a combinatorial explosion of subsequences. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Arithmetic Slices II - Subsequence?

Arithmetic Slices II - Subsequence uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: .

What are the interview tips for Arithmetic Slices II - Subsequence?

Explain your thought process clearly while coding. Consider edge cases, such as arrays with all identical elements. Discuss the trade-offs of different approaches before coding.

What are common mistakes when solving Arithmetic Slices II - Subsequence?

Failing to account for subsequences of length less than 3. Not using hashmaps effectively to store counts.

#446

Arithmetic Slices II - Subsequence

Hard

ArrayDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(n)

The optimal solution uses dynamic programming to count arithmetic subsequences efficiently. By maintaining a hashmap for each number to track the count of subsequences ending with that number and a specific difference, we avoid the combinatorial explosion of the brute-force method.

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Algorithm

4 steps

1Step 1: Initialize a list of hashmaps to store counts of subsequences for each number.
2Step 2: Iterate through each pair of numbers in the array to calculate the difference.
3Step 3: For each difference, update the hashmap for the current number based on the previous counts.
4Step 4: Sum the counts of valid subsequences that have at least three elements.

solution.py14 lines

1from collections import defaultdict
2
3def count_arithmetic_slices(nums):
4    n = len(nums)
5    dp = [defaultdict(int) for _ in range(n)]
6    count = 0
7
8    for i in range(n):
9        for j in range(i):
10            diff = nums[i] - nums[j]
11            count += dp[j][diff]
12            dp[i][diff] += dp[j][diff] + 1
13
14    return count

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we efficiently track subsequences using hashmaps, which reduces the overhead of checking each subsequence. The space complexity is O(n) because we maintain a list of hashmaps for each number.

1Using hashmaps allows for efficient counting of subsequences.
2Recognizing arithmetic sequences can be reduced to tracking differences.

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