尝试解决下面的问题复合词，无法找到返回 python 中单词的所有变体的逻辑..任何帮助将不胜感激

Question

复合词问题，这是我在下面尝试过的代码，但这不起作用，它不会返回所有可能的对，它只返回一对单词 ex:

"verylongjump": ["very", "long", "jump"])

但不会返回两者：

"verylongjump": ["very", "long", "jump"]   
"verylongjump": ["very","longjump"]

问题输入：

输入：

["long", "very", "jump", "verylongjump", "longjump", 
"iconstar", "star", "icon", "icons", "tar", "stars,"iconstars"]

预期输出：

"verylongjump": ["very", "long", "jump"]
"verylongjump": ["very","longjump"]
"iconstar": ["icon",""star"]
"iconstar":["icons", "tar"]
"iconstars":["icon","stars"]

def concatenatedWords(words):
    # Intitialize empty array for result set
    result = []

    # Helper function for Work Break II without DP but accepted by LeetCode
    def helper(idx, word, wordSet, curr, result):
        if idx == len(word):
            result.append(curr)
            return
        if idx > len(word):
            return
        for i in range(idx+1, len(word)+1):
            if word[idx:i] in wordSet:
                helper(i, word, wordSet, curr + [word[idx:I]], result)
        return

    # Convert the list of words into a HashSet
    wordSet = set(words)
    
    # Find word breaks for each word
    for word in words:
        # Initialize current word split result set
        curr = []
        helper(0, word, wordSet, [], curr)

        # Word Break solution captures the word themselves as they are present in the 
        # hash set; Filter them out by checking for length of current splits more than 1
        if len(curr) > 1:
            res = []
            # Flatten the result list
            for elem in curr:
                res.extend(elem)
        result.append(res)
    return result