## Question

Given an array `nums`

of integers, return the **length** of the longest arithmetic subsequence in `nums`

.

Recall that a *subsequence* of an array `nums`

is a list `nums[i`

with _{1}], nums[i_{2}], ..., nums[i_{k}]`0 <= i`

, and that a sequence _{1} < i_{2} < ... < i_{k} <= nums.length - 1`seq`

is *arithmetic* if `seq[i+1] - seq[i]`

are all the same value (for `0 <= i < seq.length - 1`

).

**Example 1:**

Input:nums = [3,6,9,12]Output:4Explanation:The whole array is an arithmetic sequence with steps of length = 3.

**Example 2:**

Input:nums = [9,4,7,2,10]Output:3Explanation:The longest arithmetic subsequence is [4,7,10].

**Example 3:**

Input:nums = [20,1,15,3,10,5,8]Output:4Explanation:The longest arithmetic subsequence is [20,15,10,5].

**Constraints:**

`2 <= nums.length <= 1000`

`0 <= nums[i] <= 500`

## Python Solution

class Solution: def longestArithSeqLength(self, A: List[int]) -> int: d = {} n = len(A) for i in range(n): for j in range(i+1,n): if (i,A[j]-A[i]) not in d: d[(j,A[j]-A[i])] = 1 else: d[(j,A[j]-A[i])] = 1 + d[(i,A[j]-A[i])] # print(d) m = 0 for i in d: m = max(m,d[i]) return m+1