What is the time complexity of Longest Arithmetic Subsequence of Given Difference?

The optimal solution for Longest Arithmetic Subsequence of Given Difference runs in O(n) time and uses O(n) space. This complexity is efficient because we only pass through the array once (O(n)), and we use a HashMap to store results, which allows for quick lookups.

What is the brute force solution for Longest Arithmetic Subsequence of Given Difference?

The brute force approach for Longest Arithmetic Subsequence of Given Difference is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible subsequences of the array to find the longest one that meets the arithmetic condition. This is straightforward but inefficient, as it examines every combination. The optimal Optimal Solution approach improves this to O(n).

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

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

What are the interview tips for Longest Arithmetic Subsequence of Given Difference?

Always clarify the problem requirements and constraints before jumping into coding. Think about how you can optimize a brute force solution and explain your thought process. Practice explaining your code and thought process clearly, as communication is key in interviews.

What are common mistakes when solving Longest Arithmetic Subsequence of Given Difference?

Not considering edge cases where the subsequence might only consist of one element. Failing to initialize the HashMap correctly can lead to incorrect results.

#1218

Longest Arithmetic Subsequence of Given Difference

Medium

ArrayHash TableDynamic ProgrammingHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses dynamic programming with a HashMap to track the lengths of valid subsequences ending at each number. This allows us to build on previously computed results efficiently.

⚙️

Algorithm

4 steps

1Step 1: Initialize a HashMap to store the length of the longest subsequence ending with each number.
2Step 2: Iterate through the array, and for each number, check if the number minus the difference exists in the HashMap.
3Step 3: If it exists, update the current number's length in the HashMap based on the previous number's length plus one.
4Step 4: Keep track of the maximum length found during the iteration.

solution.py7 lines

1def longest_arith_seq(arr, difference):
2    dp = {}
3    max_length = 0
4    for num in arr:
5        dp[num] = dp.get(num - difference, 0) + 1
6        max_length = max(max_length, dp[num])
7    return max_length

ℹ

Complexity note: This complexity is efficient because we only pass through the array once (O(n)), and we use a HashMap to store results, which allows for quick lookups.

1Using a HashMap allows for efficient tracking of subsequence lengths.
2Building on previously computed results is key in dynamic programming.

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