<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>

# 4.40/4.56/5.05/0/0

# nums.length <= 1000, (can run n^2 algo)
# brute force n^3 = n^3 (pick startpt, pick endpt, iterate through the list to check for progression)
# base case = 1 no
# base case = 2 no
# 3rd casd = 3 numbers, either 2 or 3 numbers 
# runtime(n^2)
# space(n*1000) -500 <= delta <= 500
# from efvery number, iterate back, dp[nums[i]-nums[i-1]]+=1
# 3, 6, 9, 12
# +3, 1

# +3,2, +6,1
# +3,2, +6,2, 

# import collections
class Solution:
    def longestArithSeqLength(self, A):
        dp = {}
        for i in range(len(A)):
            for j in range(i + 1, len(A)):
                dp[j, A[j] - A[i]] = dp.get((i, A[j] - A[i]), 1) + 1
        return max(dp.values())
        
64971