A Trie, often pronounced as "try," is a specialized tree-like data structure that is primarily used to store a dynamic set or associative array of strings. The main properties of a trie include:
Consider the following set of words: ["cat", "dog", "car", "bat"]
. The representing trie would look like this:
(root)
/ | \
c d b
/ \ | \
a a o a
| | | |
t r g t
Here:
Dynamic programming is an optimization technique used to solve complex problems by breaking them down into simpler subproblems. It efficiently stores the results of these subproblems to avoid redundant calculations, yielding a significant performance improvement over naive solutions.
Combining tries with dynamic programming is particularly useful in string-related problems, such as searching for valid words in a dictionary or finding the longest subsequence of a string that satisfies certain criteria. The trie structure allows for efficient prefix retrieval, while dynamic programming helps in systematically building up solutions to problems.
Let’s consider a practical example: Finding the number of ways to segment a given string into words from a dictionary.
Given a string s
and a list of words (dictionary), determine the total number of ways to break s
into valid words.
s = "applepie" words = ["apple", "pie", "app"]
dp
of size n+1
where n
is the length of the string s
. Initialize dp[0] = 1
as there is one way to segment an empty string.i
in the string, use the trie to check for valid words ending at that index.Here’s how you can implement this:
class TrieNode: def __init__(self): self.children = {} self.is_end_of_word = False class Trie: def __init__(self): self.root = TrieNode() def insert(self, word): node = self.root for char in word: if char not in node.children: node.children[char] = TrieNode() node = node.children[char] node.is_end_of_word = True def count_segmentations(s, words): trie = Trie() for word in words: trie.insert(word) n = len(s) dp = [0] * (n + 1) dp[0] = 1 # Base case for i in range(1, n + 1): node = trie.root for j in range(i - 1, -1, -1): char = s[j] if char in node.children: node = node.children[char] if node.is_end_of_word: dp[i] += dp[j] else: break return dp[n] # Example Usage s = "applepie" words = ["apple", "pie", "app"] print(count_segmentations(s, words))
Trie
class creates a trie structure where each word from the dictionary is inserted.m
is the total characters across all words in the dictionary.With this approach, you're effectively utilizing tries to optimize the prefix lookups while leveraging dynamic programming to accumulate results, resulting in an elegant solution to string segmentation problems.
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