Dynamic Programming, create a memo table longest stable subsequence - python-3.x

Question

I have worked on a dynamic programming problem for quite some time now and am stuck, so any help is much appreciated.
Here is the first part of the problem which I was able to get the tests to pass.
def lssLength(a, i, j):
aj = a[j] if 0 <= j < len(a) else None
# Implement the recurrence below. Use recursive calls back to lssLength
assert 0 <= i <= len(a)
if i >= len(a) or j >= len(a):
return 0
if aj and abs(a[i] - a[j]) > 1:
return lssLength(a, i+1, j)
if aj is None or (abs(a[i] - a[j]) <= 1 and i != j):
return max(lssLength(a, i+1, j), lssLength(a, i+1, i) + 1)
else:
return lssLength(a, i+1, j)
Here are my test cases for the first problem:
def test_lss_length(self):
# test 1
n1 = lssLength([1, 4, 2, -2, 0, -1, 2, 3], 0, -1)
print(n1)
self.assertEqual(4, n1)
# test 2
n2 = lssLength([1, 2, 3, 4, 0, 1, -1, -2, -3, -4, 5, -5, -6], 0, -1)
print(n2)
self.assertEqual(8, n2)
# test 3
n3 = lssLength([0, 2, 4, 6, 8, 10, 12], 0, -1)
print(n3)
self.assertEqual(1, n3)
# test 4
n4 = lssLength(
[4, 8, 7, 5, 3, 2, 5, 6, 7, 1, 3, -1, 0, -2, -3, 0, 1, 2, 1, 3, 1, 0, -1, 2, 4, 5, 0, 2, -3, -9, -4, -2, -3,
-1], 0, -1)
print(n4)
self.assertEqual(14, n4)
Now I need to take the recursive solution and convert it to dynamic programming, and this is where I am stuck. I am using the same tests as before, but the tests are failing.
def memoizeLSS(a):
T = {} # Initialize the memo table to empty dictionary
# Now populate the entries for the base case
# 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)]
n = len(a)
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):
if abs(a[i] - a[j]) > 1:
try:
T[(i, j)] = max(0, T[(i, j+1)], T[(i+1, j)])
except Exception:
T[(i, j)] = 0
elif abs(a[i] - a[j]) <= 1 and i != j:
T[(i, j)] = T[(i+1, j+1)] + 1
else:
T[(i, j)] = max(0, T[(i+1, j+1)])
for i in range(n-2, -1, -1):
T[(i, -1)] = max(T[(i+1, -1)], T[(i+1, 0)], T[(i, 0)], 0)
return T
If you've read all of this, thank you so much. I know it is a lot and really appreciate your time. Any pointers to reading materials, etc. is much appreciated.
If there are more details required, please let me know. Thanks.

Answer 1

Your solution only worked for the first test case. Below is a corrected version:
def memoizeLSS(a):
T = {} # Initialize the memo table to empty dictionary
n = len(a)
for j in range(-1, n):
T[(n, j)] = 0 # i = n and j
# Now populate the entries for the base case
# 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)]
n = len(a)
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] - a[j]) > 1:
T[(i, j)] = T[(i+1, j)]
elif aj == None or abs(a[i] - a[j]) <= 1:
T[(i, j)] = max(T[(i+1, i)] + 1, 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)
return T