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This approach involves using a Trie data structure to store all words. We then perform a DFS to check if a word can be formed using other words in the trie. The main advantage of using a Trie is that it provides an efficient way to store and look up words simply by traversing nodes based on each character of the word.
We will mark nodes in the Trie that represent the end of a word and during DFS checks if the current segment of the word is in the Trie.
Time Complexity: O(n * m^2), where n is the number of words and m is the average length of the words. The DFS function can make at most m calls for each word.
Space Complexity: O(n * m) for storing the Trie, where n is the number of words and m is the average length of the words.
1class TrieNode:
2 def __init__(self):
3 self.children = {}
4 self.is_end_of_word = False
5
6class Solution:
7 def insert(self, root, word):
8 node = root
9 for char in word:
10 if char not in node.children:
11 node.children[char] = TrieNode()
12 node = node.children[char]
13 node.is_end_of_word = True
14
15 def dfs(self, root, word, start, count):
16 if start == len(word):
17 return count >= 2
18
19 node = root
20 for i in range(start, len(word)):
21 char = word[i]
22 if char not in node.children:
23 return False
24 node = node.children[char]
25 if node.is_end_of_word:
26 if self.dfs(root, word, i + 1, count + 1):
27 return True
28 return False
29
30 def findAllConcatenatedWordsInADict(self, words):
31 root = TrieNode()
32 for word in words:
33 if word:
34 self.insert(root, word)
35
36 result = []
37 for word in words:
38 if self.dfs(root, word, 0, 0):
39 result.append(word)
40 return result
41
42# Usage example
43if __name__ == "__main__":
44 sol = Solution()
45 words = ["cat", "cats", "catsdogcats", "dog", "dogcatsdog", "hippopotamuses", "rat", "ratcatdogcat"]
46 concatenated_words = sol.findAllConcatenatedWordsInADict(words)
47 print("Concatenated Words:")
48 for word in concatenated_words:
49 print(word)
50In this Python solution, a simple TrieNode class is defined to construct the Trie, and a solution class inserts each word into the Trie. Using depth-first search (DFS), it examines potential concatenations by checking if sub-words exist in the Trie.
In this approach, dynamic programming (DP) is used to efficiently solve the problem of finding concatenated words. The idea is to treat each word as a state and store the results of subproblems to avoid repeated calculations. We iterate over each word and for each segment of the word, check if it can be split into valid sub-words based on previously computed results in the DP array.
This can be visualized as using a DP boolean array where each index represents if the word up to that index can be segmented into valid words in the list. This method is beneficial in reducing the number of redundant calculations.
Time Complexity: O(n * m^2), where n is the number of words and m is the average length of the words. The space used for the dp array and hash table is the main contributor.
Space Complexity: O(n * m) due to hash table entry needs.
The JavaScript solution revolves around dynamic programming with a Set for word storage. The main function iteratively evaluates if any word can be assembled from combinations of shorter words, tracked by results in a defined boolean array 'dp'. The subsequences yielding potential matches mark potential composite structures, fully utilizing JS's substring evaluation.