<p>Given an array <code>nums</code> of integers, return <em>the length of the longest arithmetic subsequence in</em> <code>nums</code>.</p>

<p><strong>Note</strong> that:</p>

<ul>
	<li>A <strong>subsequence</strong> is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.</li>
	<li>A sequence <code>seq</code> is arithmetic if <code>seq[i + 1] - seq[i]</code> are all the same value (for <code>0 &lt;= i &lt; seq.length - 1</code>).</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 500</code></li>
</ul>

class Solution:
    def longestArithSeqLength(self, nums: List[int]) -> int:
        #brute force
        #find the longest subsequence for every i and for every jump value
        #n^2 * m 
        #what if we go backwards and treat this as a recursive prob
        #start iwth 10, len1
        #2, 10: 2 with diff 8
        #7 2 10: 2 with diff 3 and 8
        #4 7 2 10: 4 -> 7 has difff 3 so max len is now 3
        #what if I cache the val and diff in a map
        #and for each node I search -500 -> 500 to see if nums[i] + j, j is a previously seen node
        #space is O(n^m) runtime is O*(n*m)
        #idt this problem is possible in O(n) or O(m) maybe O(n+m)?
        
        #ok lets try the dp soln

        opt = defaultdict(int)
        seen = set()
        ans = 1
        m = max(nums)
        for i in range(len(nums)):
            for j in range(-m, m+1):
                if (nums[i] - j) in seen:
                    opt[(nums[i], j)] = max(2, 1 + opt[(nums[i] - j, j)])
                    ans = max(ans, opt[(nums[i], j)])

                
            seen.add(nums[i])
        return ans
54296