DP Solution
IntuitionThe goal of the problem is to find the longest subsequence in an array such that the absolute differences between consecutive elements form a non-incre
# Intuition
The goal of the problem is to find the longest subsequence in an array such that the absolute differences between consecutive elements form a non-increasing sequence. To achieve this, we can leverage dynamic programming to track the longest subsequence for each possible value of the difference between consecutive elements.

We need to consider two things:
1. The current number in the subsequence.
2. The difference between consecutive elements in the subsequence.

By maintaining a dynamic programming table, we can efficiently compute the length of the longest valid subsequence that adheres to the problem constraints.

# Approach
We'll use dynamic programming (DP) to solve the problem. The idea is to maintain a DP table, where `dp[i][d]` represents the longest subsequence that ends with element `i` and has an absolute difference `d` between consecutive elements. The main steps of the approach are as follows:

1. **Initialize the DP table**: We will initialize a table of size 302 x 302, where 302 is enough to handle the range of differences. `dp[i][d]` will store the length of the longest subsequence ending at element `i` with a difference `d`.

2. **Iterate through the array**: For each element in the array `nums`, calculate all possible differences with the next element in the array. Update the DP table by considering whether adding the current element to an existing subsequence with the given difference `d` forms a valid subsequence.

3. **Final Answer**: After processing all elements, the result will be the maximum value in the DP table, which gives the length of the longest valid subsequence.

# Complexity
- **Time complexity**: $$O(n \times 300)$$, where `n` is the length of the array. This is because we process each element, and for each element, we update the DP table for possible differences between 1 and 300.
  
- **Space complexity**: $$O(300 \times 300) = O(1)$$, since the DP table size is fixed at 302 x 302.

# Code
```java
class Solution {
    public int longestSubsequence(int[] nums) {
        int[][] dp = new int[302][302];  // DP table

        // Iterate from the last element to the first
        for (int i = nums.length - 1; i >= 0; --i) {
            int num = nums[i];
    
            // Iterate through all possible next elements (1 to 300)
            for (int next = 1; next <= 300; ++next) {
                int diff = Math.abs(next - num);  // Calculate the absolute difference
                dp[num][diff] = Math.max(dp[num][diff], dp[next][diff] + 1);  // Update DP table
            }

            // Ensure the DP table is non-decreasing by difference
            for (int j = 1; j <= 300; ++j) {
                dp[num][j] = Math.max(dp[num][j], dp[num][j - 1]);
            }
        }

        // Find the maximum value in the DP table
        int ans = Integer.MIN_VALUE;
        for (int i = 0; i <= 301; ++i) {
            for (int j = 0; j <= 301; ++j) {
                ans = Math.max(ans, dp[i][j]);  // Maximize the result
            }
        }

        return ans;
    }
}
```

This solution efficiently computes the longest subsequence based on the problem constraints. By using dynamic programming and carefully updating the DP table, we ensure that the sequence formed follows the required condition that the absolute differences between consecutive elements are non-increasing.
['Dynamic Programming', 'Java']