Simple Dynamic Programming Approach
IntuitionThe problem asks for the longest subsequence in an array where the absolute difference between consecutive elements is the same. We want to find such a
# Intuition
The problem asks for the longest subsequence in an array where the absolute difference between consecutive elements is the same. We want to find such a subsequence that satisfies this condition and has the longest length. This type of problem is a variation of finding the longest arithmetic subsequence in a list.

# Approach
The approach uses dynamic programming with memoization to track the longest subsequence for each possible common difference d between elements in the array. The idea is to maintain a dp array where dp[i][d] represents the length of the longest subsequence that ends at index i with a common difference d. Additionally, the mx array is used to store the maximum subsequence length at each index for each difference, which allows for efficient updates during iteration.

1) Initialization: We start by initializing dp and mx arrays with values of 1, since a subsequence with a single element is trivially of length 1.

2) Updating dp and mx:
i) For each pair of indices i and j (where j < i), we calculate the absolute difference d between nums[i] and nums[j].
ii) Using the mx array, we update the dp[i][d] value by checking if we can extend a subsequence ending at j with difference d.
iii) After processing each index i, we update the mx[i][d] to store the maximum length of any subsequence ending at i for each difference.

3) Final Answer: The longest subsequence is stored in the variable ans, which is updated during each iteration.

# Code
```cpp []
class Solution {
public:
    int longestSubsequence(vector<int>& nums) {
        int n = nums.size(), ans = 1;
        static int dp[10001][301], mx[10001][301];

        // Initializing dp and mx arrays
        for (int i = 0; i < n; i++) {
            for (int d = 0; d <= 300; d++) {
                dp[i][d] = 1;
                mx[i][d] = 1;
            }
        }

        // Process the numbers
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                int d = abs(nums[i] - nums[j]);
                dp[i][d] = max(dp[i][d], mx[j][d] + 1);
                ans = max(ans, dp[i][d]);
            }

            // Update the mx array
            for (int d = 300; d >= 0; d--) {
                mx[i][d] = max(mx[i][d + 1], dp[i][d]);
            }
        }

        return ans;
    }
};
```
['C++']