How do you solve Longest Arithmetic Subsequence on LeetCode?

Longest Arithmetic Subsequence (LeetCode #1027) 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: Understanding the difference between subsequences and subsequences with specific properties is crucial.

What is the brute force solution for Longest Arithmetic Subsequence?

The brute force approach for Longest Arithmetic Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subsequence of the array to see if it forms an arithmetic sequence. This is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Longest Arithmetic Subsequence?

Longest Arithmetic Subsequence uses the following concepts: Array, Hash Table, Binary Search, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Arithmetic Subsequence?

Clearly explain your thought process and the reasoning behind each step. Be prepared to optimize your solution if the interviewer asks for it. Practice identifying patterns in problems, such as arithmetic sequences.

What are common mistakes when solving Longest Arithmetic Subsequence?

Not considering all pairs when calculating differences. Confusing subsequences with contiguous subarrays.

#1027

Longest Arithmetic Subsequence

Medium

ArrayHash TableBinary SearchDynamic ProgrammingHash MapArray

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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 track the longest arithmetic subsequence ending at each index. We leverage the differences between pairs of elements to build our solution efficiently.

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Algorithm

3 steps

1Step 1: Create a 2D array (or a hashmap) to store lengths of arithmetic subsequences with a specific difference.
2Step 2: Iterate through each pair of elements in the array to calculate the difference.
3Step 3: Update the length of the arithmetic subsequence ending at the current index based on previously computed lengths.

solution.py18 lines

1# Full working Python code
2from collections import defaultdict
3
4def longestArithSeqLength(nums):
5    n = len(nums)
6    if n < 2:
7        return n
8    dp = defaultdict(lambda: defaultdict(int))
9    max_length = 2
10
11    for j in range(n):
12        for i in range(j):
13            diff = nums[j] - nums[i]
14            dp[diff][j] = dp[diff][i] + 1
15            max_length = max(max_length, dp[diff][j] + 1)
16
17    return max_length
18

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loop structure, but we use additional space for the hashmap to store lengths of subsequences, leading to a space complexity of O(n).

1Understanding the difference between subsequences and subsequences with specific properties is crucial.
2Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.

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