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# Longest Arithmetic Sequence

Given an array `A` of integers, return the **length** of the longest arithmetic subsequence in `A`.

Recall that a *subsequence* of `A` is a list `A[i_1], A[i_2], ..., A[i_k]` with `0 <= i_1 < i_2 < ... < i_k <= A.length - 1`, and that a sequence `B` is *arithmetic* if `B[i+1] - B[i]` are all the same value (for `0 <= i < B.length - 1`).

**Example 1:**

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

**Example 2:**

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

**Example 3:**

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

**Note:**

1. `2 <= A.length <= 2000`
2. `0 <= A[i] <= 10000`

```java
class Solution {
    // The main idea is to maintain a map of differences seen at each index.
    public int longestArithSeqLength(int[] A) {
        if (A.length <= 1)
            return A.length;
        int longest = 0;
        HashMap<Integer, Integer>[] dp = new HashMap[A.length];
        for (int i = 0; i < A.length; ++i) {
            // Initialize the map.
            dp[i] = new HashMap<Integer, Integer>();
        }
        for (int i = 1; i < A.length; ++i) {
            int x = A[i];
            for (int j = 0; j < i; ++j) {
                int y = A[j];
                int d = x - y;
                // We at least have a minimum chain length of 2 now,
                // given that (A[j], A[i]) with the difference d,
                // by default forms a chain of length 2.
                int len = 2;
                if (dp[j].containsKey(d)) {
                    // At index j, if we had already seen a difference d,
                    // then potentially, we can add A[i] to the same chain
                    // and extend it by length 1.
                    len = dp[j].get(d) + 1;
                }
                // Update the max chain length for difference d at index i.
                dp[i].put(d, len);
                // Update the global max.
                longest = Math.max(longest, dp[i].get(d));
            }
        }
        return longest;
    }
}
```


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