Longest arithmetic subsequence of given difference

Time: O(N); Space: O(N); medium

Given an integer array arr and an integer difference, return the length of the longest subsequence in arr which is an arithmetic sequence such that the difference between adjacent elements in the subsequence equals difference.

Example 1:

Input: arr = [1,2,3,4], difference = 1

Output: 4

Explanation:

  • The longest arithmetic subsequence is [1,2,3,4].

Example 2:

Input: arr = [1,3,5,7], difference = 1

Output: 1

Explanation:

  • The longest arithmetic subsequence is any single element.

Example 3:

Input: arr = [1,5,7,8,5,3,4,2,1], difference = -2

Output: 4

Explanation:

  • The longest arithmetic subsequence is [7,5,3,1].

Constraints:

  • 1 <= len(arr) <= 10^5

  • -10^4 <= arr[i], difference <= 10^4

Hints:

  1. Use dynamic programming.

  2. Let dp[i] be the maximum length of a subsequence of the given difference whose last element is i.

  3. dp[i] = 1 + dp[i-k]

1. Dynamic programming [O(N), O(N)]

[1]:

import collections

class Solution1(object):
    """
    Time: O(N)
    Space: O(N)
    """
    def longestSubsequence(self, arr, difference):
        """
        :type arr: List[int]
        :type difference: int
        :rtype: int
        """
        result = 1
        lookup = collections.defaultdict(int)

        for i in range(len(arr)):
            lookup[arr[i]] = lookup[arr[i]-difference] + 1
            result = max(result, lookup[arr[i]])

        return result

[2]:

s = Solution1()

arr = [1,2,3,4]
difference = 1
assert s.longestSubsequence(arr, difference) == 4

arr = [1,3,5,7]
difference = 1
assert s.longestSubsequence(arr, difference) == 1

arr = [1,5,7,8,5,3,4,2,1]
difference = -2
assert s.longestSubsequence(arr, difference) == 4