I have written two versions of Word Break exhaustive combinations. Please comment

I am solving Word Break II from leetcode.

https://leetcode.com/problems/word-break-ii/description/

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

Note:

The same word in the dictionary may be reused multiple times in the segmentation. You may assume the dictionary does not contain duplicate words.

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
[
"pine apple pen apple",
"pineapple pen apple",
"pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.

I think my Python has improved since I posted island problem last. Your input is appreciated.

Here are the two solutions:

def inverse_iterative(s, wdict):
st = []
ans = []

for w in wdict:
if s.startswith(w):
if len(s) == len(w):
ans.append(w)
return ans
st.append((w, 0, [w]))

while st:
cur_w, cur_idx, cur_list = st.pop()
cur_start = cur_idx + len(cur_w)

for w in wdict:
if not s[cur_start:].startswith(w):
continue
if len(w) == len(s[cur_start:]):
ans.append(' '.join(cur_list+[w]))
else:
st.append((w, cur_start, cur_list+[w]))
return ans

def inverse_helper(s, wdict, memo):
if s in memo:
return memo[s]
if not s:
return []

res = []
for word in wdict:
if not s.startswith(word):
continue
if len(word) == len(s):
res.append(word)
else:
result_of_rest = helper(s[len(word):], wordDict, memo)
for r in result_of_rest:
r = word + ' ' + r
res.append(r)

memo[s] = res
return res

ans_iterative = inverse_iterative(s, wdict)
ans_recursive = inverse_helper(s, wdict, {})