Fastest JAVA solution. Beats 100% in TC and SC.
IntuitionSee how the hints 1 and 2 are given for question. Once we understand that the we just need to figure out the answer for each index and difference it be
# Intuition
See how the hints 1 and 2 are given for question. Once we understand that the we just need to figure out the answer for each index and difference it becomes easy to understand.

# Approach
Initialization:

Create a 2D DP table, dp[301][301], where dp[g][d] represents the longest subsequence that ends with the value g and difference d between consecutive elements. The size of 301 is chosen because the values of nums[i] are between 1 and 300, and the differences between consecutive elements also range from 0 to 300.
Initially, set all values in the DP table to 0.
First Step:

For the first number in the array, initialize all possible differences (from 0 to 300) as a subsequence of length 1, because any element can start a subsequence by itself.
Main Logic:

For each subsequent element in the array:
Try to extend the subsequences ending at previous elements, considering both increasing and decreasing differences.
For each possible difference, update the DP table to reflect the longest subsequence possible that ends at the current element.
After updating the DP table for each element, propagate the longest subsequences backward to maintain the non-increasing difference condition.
Result:

The result is the maximum value found in the DP table, which represents the length of the longest subsequence where the absolute differences between consecutive elements are non-increasing.

# Complexity
- Time complexity: O(300 * n) ~ O(kn) 

- Space complexity: O(300 * 300) ~ O(kn)

# Code
```java []
class Solution {
    public int longestSubsequence(int[] nums) {
        int[][] dp=new int[301][301];
        int max=0;
        for(int i=0;i<=300;i++)dp[nums[0]][i]=1;
        for(int i=1;i<nums.length;i++){
            int g=nums[i];
            for(int j=0;j<=300;j++){
                int curr=0;
             if(g-j>0){
                curr=Math.max(dp[g-j][j]+1,curr);
             }
             if(g+j<=300){
               curr=Math.max(dp[g+j][j]+1,curr);
             }
             dp[g][j]=Math.max(curr,dp[g][j]);
            }
            for(int j=299;j>=0;j--){
               dp[g][j]=Math.max(dp[g][j],dp[g][j+1]);
               max=Math.max(max,dp[g][j]);
            }
            
        }
        return max;
        
    }

    
}

```
['Math', 'Dynamic Programming', 'Prefix Sum', 'Java']