Longest Stable Subsequence - Recover solution from memo table

I am working on a DP problem to find longest stable subsequence, and I'm currently stuck with recovering the solution.

Here is the problem statement

I have tried the below solution,

def computeLSS(a):
    T = {} # Initialize the memo table to empty dictionary
    # Now populate the entries for the base case 
    n = len(a)
    for j in range(-1, n):
        T[(n, j)] = 0 # i = n and j 
    # Now fill out the table : figure out the two nested for loops
    # It is important to also figure out the order in which you iterate the indices i and j
    # Use the recurrence structure itself as a guide: see for instance that T[(i,j)] will depend on T[(i+1, j)]
    # your code here
    
    for i in range(0, n + 1):
        for j in range(-1, n + 1):
            T[(i, j)] = 0
    
    for i in range(n - 1, -1, -1):
        for j in range(n - 1 , -1, -1):
            aj = a[j] if 0 <= j < len(a) else None 
            if aj != None and abs(a[i] - aj) > 1:
                T[(i, j)] = T[(i + 1, j)]
            if aj == None or abs(a[i] - aj) <= 1:
                T[(i, j)] = max(1 + T[(i + 1), i], T[(i + 1, j)])
    
    for i in range(n-2, -1, -1):
        T[(i, -1)] = max(T[(i+1, -1)], T[(i+1, 0)], T[(i, 0)], 0)
                
    
    i = 0
    j = -1
    sol = []
    while i < n and j < n:
        if abs(T[(i, j)] - T[(i+1, j)]) > 1:
            sol.append(a[i])
            j = i
        i = i + 1
        
    return sol

But it fails for the below test cases,

a2 = [1, 2, 3, 4, 0, 1, -1, -2, -3, -4, 5, -5, -6]
print(a2)
sub2 = computeLSS(a2)
print(f'sub2 = {sub2}')
assert len(sub2) == 8

a3 = [0,2, 4, 6, 8, 10, 12]
print(a3)
sub3 = computeLSS(a3)
print(f'sub3 = {sub3}')
assert len(sub3) == 1

I would really appreciate if someone can help me with some pointers like what might be the problem with my recovery code?