Problem Statement

Given an array nums containing positive integers, return the maximum length of a subsequence that forms an arithmetic progression.

• A subsequence is an array that can be formed from nums by deleting 0 or more elements without changing the order of the remaining elements.

• An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive elements is constant. In short, seq[i + 1] - seq[i] should be same for all 0 <= i < seq.length - 2.

Examples

• Example 1:

  • Input: [8, 12, 6, 4, 2]
  • Expected Output: 4
  • Justification: The subsequence [8, 6, 4, 2] forms the longest arithmetic subsequence with a common difference of -2, thus the longest arithmetic subsequence has a length 4.

• Example 2:

  • Input: [5, 10, 15, 7, 11, 5]
  • ****Expected Output: 3
  • Justification:** The subsequence [5, 10, 15] forms the longest arithmetic subsequence with a common difference of 5.

• Example 3:

  • Input: [1, 7, 10, 15, 27, 29]
  • Expected Output: 3
  • Justification: The subsequence [1, 15, 29] forms the longest arithmetic subsequence with a common difference of 14.

Solution

To solve this problem, a dynamic programming approach is most effective. The key idea is to use a hash table to store the lengths of the longest arithmetic subsequences ending at each element, indexed by their differences. This method allows for an efficient update of subsequences as the array is iterated through.

By considering each pair of elements in the array, we can update the hash table entries to reflect the longest subsequence lengths observed so far. This approach is favored because it enables the algorithm to consider all possible pairs of elements and their potential to extend existing arithmetic subsequences without needing to explicitly track each subsequence.

Step-by-step Algorithm

• Initialize a list of hash maps, where each hash map corresponds to an element in the input array. Each map will track the lengths of arithmetic subsequences ending with that element, keyed by their common differences.
• Iterate through each pair of elements in the array. For each pair (i, j) where i < j, calculate the difference diff = array[i] - array[j].
• For each difference, update the hash map corresponding to array[i] to reflect the longest subsequence ending at i with difference diff. This is done by checking the length of the longest subsequence ending at j with the same difference and incrementing it by one.
• If no such subsequence exists for array[j], initialize it with a length of 2 (since at least two elements are needed to form a subsequence).
• Keep track of the maximum length found during this process.
• Return the maximum length as the result.

Algorithm Walkthrough

Consider the input [1, 7, 10, 15, 27, 29].

Initialization

• Initialize dp as an array of hash maps, one for each element in A.
• result is initialized to 2, the minimum possible length of an arithmetic sequence.

Processing

Iteration i = 1 (Element 7):

• j = 0: Compare 7 with 1.
  • Difference diff = 6. Update dp[1][6] = 2.
  • HashMap for i=1: {6: 2}.
• maxLength: 2

Iteration i = 2 (Element 10):

• j = 0: Compare 10 with 1.

  • Difference diff = 9. Update dp[2][9] = 2.
  • HashMap for i=2 after j=0: {9: 2}.

• j = 1: Compare 10 with 7.

  • Difference diff = 3. Update dp[2][3] = 2.
  • HashMap for i=2 after j=1: {9: 2, 3: 2}.

• maxLength: 2

Iteration i = 3 (Element 15):

• j = 0: Compare 15 with 1.
  • Difference diff = 14. Update dp[3][14] = 2.
  • HashMap for i=3 after j=0: {14: 2}.
• j = 1: Compare 15 with 7.
  • Difference diff = 8. Update dp[3][8] = 2.
  • HashMap for i=3 after j=1: {14: 2, 8: 2}.
• j = 2: Compare 15 with 10.
  • Difference diff = 5. Update dp[3][5] = 2.
  • HashMap for i=3 after j=2: {14: 2, 8: 2, 5: 2}.
• maxLength: 2

Iteration i = 4 (Element 27):

• j = 0: Compare 27 with 1.
  • Difference diff = 26. Update dp[4][26] = 2.
  • HashMap for i=4 after j=0: {26: 2}.
• j = 1: Compare 27 with 7.
  • Difference diff = 20. Update dp[4][20] = 2.
  • HashMap for i=4 after j=1: {26: 2, 20: 2}.
• j = 2: Compare 27 with 10.
  • Difference diff = 17. Update dp[4][17] = 2.
  • HashMap for i=4 after j=2: {26: 2, 20: 2, 17: 2}.
• j = 3: Compare 27 with 15.
  • Difference diff = 12. Update dp[4][12] = 2.
  • HashMap for i=4 after j=3: {26: 2, 20: 2, 17: 2, 12: 2}.
• maxLength: 2

Iteration i = 5 (Element 29):

• j = 0: Compare 29 with 1.

  • Difference diff = 28. Update dp[5][28] = 2.
  • HashMap for i=5 after j=0: {28: 2}.

• j = 1: Compare 29 with 7.

  • Difference diff = 22. Update dp[5][22] = 2.
  • HashMap for i=5 after j=1: {28: 2, 22: 2}.

• j = 2: Compare 29 with 10.

  • Difference diff = 19. Update dp[5][19] = 2.
  • HashMap for i=5 after j=2: {28: 2, 22: 2, 19: 2}.

• j = 3: Compare 29 with 15.

  • Difference diff = 14. Update dp[5][14] = dp[3].getOrDefault(14, 1) + 1 = 3.
  • This update extends the sequence [1, 15] to [1, 15, 29], with a common difference of 14.
  • HashMap for i=5 after j=3: {28: 2, 22: 2, 19: 2, 14: 3}.

• j = 4: Compare 29 with 27.

  • Difference diff = 2. Update dp[5][2] = 2.
  • HashMap for i=5 after j=4: {28: 2, 22: 2, 19: 2, 14: 3, 2: 2}.

• maxLength: 3

• The final result is updated to 3 when processing i = 5 and j = 3, identifying the longest arithmetic subsequence [1, 15, 29] with a difference of 14.

Code

Complexity Analysis

Time Complexity

• O(n^2): The primary operation in the algorithm is a double loop over the input array of size (n), where (n) is the length of the input array. For each element, the algorithm calculates the difference with every other element that comes before it, updating the hash map accordingly. This nested iteration leads to a quadratic time complexity.

Space Complexity

• O(n^2): The algorithm maintains a hash map for each of the (n) elements in the input array, where each hash map can potentially store up to (n-1) key-value pairs (representing the arithmetic differences and their corresponding lengths). This results in a space complexity that is quadratic in the worst case.