Longest Arithmetic Subsequence

class Solution {
    public int longestArithmeticSubsequence(int[] array) {
        int result = 0;
        HashMap<Integer, Integer>[] sequenceLength = new HashMap[array.length];
        for (int i = 0; i < array.length; ++i) {
            sequenceLength[i] = new HashMap<>();
            for (int j = 0; j < i; ++j) {
                int difference = array[j] - array[i];
                sequenceLength[i].put(difference, sequenceLength[j].getOrDefault(difference, 1) + 1);
                result = Math.max(result, sequenceLength[i].get(difference));
            }
        }
        return result;
    }
}

This summary guides you through implementing a solution to find the longest arithmetic subsequence within an array using Java. The approach leverages dynamic programming concepts with HashMaps for efficient computation.

• Initiate result to store the maximum length of the found subsequences.
• Create an array sequenceLength of HashMaps, where each map stores the length of the arithmetic subsequences using common differences as keys.
• Iterate through each element i of the given array:
  • Initialize sequenceLength[i] as a new HashMap.
  • Iterate through previous elements j (from 0 to i-1):
    • Calculate the difference between the current element array[i] and array[j].
    • Update the map for sequenceLength[i], setting the difference as the key. If the key exists in sequenceLength[j], increment its value by 1; otherwise, start from 2 (since a difference involving two numbers denotes the start of a potential sequence).
    • Update the result if the current subsequence length is greater than the previously recorded maximum.
• Return result, which now holds the length of the longest arithmetic subsequence found.

This solution efficiently records possible subsequences by their differences and lengths, avoiding repetitive checks and maximizing computational efficiency through immediate access via map keys.

class Solution:
    def maxArithmeticLength(self, values: List[int]) -> int:
        arithmetic_sequences = {}

        for j in range(len(values)):
            for i in range(j):
                diff = values[j] - values[i]
                key = (j, diff)
                if (i, diff) in arithmetic_sequences:
                    arithmetic_sequences[key] = arithmetic_sequences[(i, diff)] + 1
                else:
                    arithmetic_sequences[key] = 2

        return max(arithmetic_sequences.values())

The Python3 solution provided is designed to find the length of the longest arithmetic subsequence in a given list values. Here’s a breakdown of how the solution operates:

1. Initializes a dictionary arithmetic_sequences to keep track of possible arithmetic sequences and their lengths.

2. Utilizes two nested loops where the outer loop variable j represents the current end of the subsequence, and the inner loop variable i represents the potential start of the subsequence. This explores every possible subsequence pair in the list.

3. Calculates diff as the difference between the elements at the j and i indices, which serves as the common difference of the arithmetic sequence starting at index i and ending at index j.

4. Constructs a key tuple consisting of the current index j and the diff, representing a unique identifier for that particular arithmetic sequence in regard to its ending index and common difference.

5. Checks if there exists an entry for the sequence with one less element (ending at index i with the same common difference). If it exists, it extends this sequence by adding the current element at index j, otherwise, it starts a new sequence including only elements at indices i and j.

6. Upon completion of the loops, returns the length of the longest arithmetic subsequence by finding the maximum value in the arithmetic_sequences dictionary.

This approach is efficient as it avoids recalculating sequences from scratch by dynamically storing and updating lengths of identified arithmetic sequences.