💡 Problem Formulation: The task at hand involves determining whether a given string can be broken down into a sequence of words that are present within a specified list. For instance, if the input string is “applepie” and the list of words contains [“apple”, “pie”], the desired output should be True since “applepie” can be split into “apple” and “pie”, both of which are in the list.
Method 1: Recursive Approach with Memoization
This method involves a recursive function that checks if the string can be segmented by trying to match the prefix of the string with the words in the list, and then recursing on the remainder of the string. Memoization is used to store already computed results to avoid redundant calculations and hence optimize the process.
Here’s an example:
def canBreak(s, wordDict, memo={}):
if s in memo: return memo[s]
if s == "": return True
for word in wordDict:
if s.startswith(word):
if canBreak(s[len(word):], wordDict, memo):
memo[s] = True
return True
memo[s] = False
return False
word_list = ["apple", "pie", "banana"]
string_check = "applepie"
print(canBreak(string_check, word_list))Output: True
The code snippet defines a function canBreak that takes a string s, a list of words wordDict, and a memoization dictionary memo. It checks if the string starts with any word from the list; if it does, it recurses on the remainder of the string. The result of each call is stored in memo to prevent duplicate work.
Method 2: Dynamic Programming
Dynamic programming can efficiently solve this problem by creating a boolean array dp where each element at index i is True if the substring up to that index can be segmented into the list of words. The algorithm iterates over the string, updating the dp array based on the previous computations.
Here’s an example:
def canBreak(s, wordDict):
dp = [False] * (len(s) + 1)
dp[0] = True
for i in range(1, len(s) + 1):
for word in wordDict:
if dp[i - len(word)] and s[i - len(word):i] == word:
dp[i] = True
return dp[len(s)]
word_list = ["apple", "pie", "banana"]
string_check = "applepie"
print(canBreak(string_check, word_list))Output: True
The function canBreak utilizes a dp array where each element indicates whether the substring up to that point can be decomposed. It scans the string left to right, using previously computed values to build up to the final result, which indicates whether the entire string can be segmented into words from the list.
Method 3: Breadth-First Search
By interpreting the problem as a graph search, we can apply Breadth-First Search (BFS) to find if there’s a path from the beginning to the end of the string. Each node represents the starting index of an unused portion of the string, and an edge connects to a node if there’s a word in the list matching the beginning of this unused portion.
Here’s an example:
from collections import deque
def canBreak(s, wordDict):
word_set = set(wordDict)
q = deque([0])
seen = set()
while q:
start = q.popleft()
if start in seen:
continue
seen.add(start)
for end in range(start + 1, len(s) + 1):
if s[start:end] in word_set:
if end == len(s):
return True
q.append(end)
return False
word_list = ["apple", "pie", "banana"]
string_check = "applepie"
print(canBreak(string_check, word_list))Output: True
The provided code utilizes BFS to explore the string by maintaining a queue of indices q. For each starting index, it attempts to find a matching word in the word_set. If a match spans the full string length, True is returned; otherwise, the search continues.
Method 4: Using Trie Data Structure
The Trie data structure is well-suited for this problem since it allows for a fast and efficient search for prefixes matching the words in the list. Building a trie out of the list of words, we can navigate through it while scanning the string to check if it can be segmented.
Here’s an example:
class TrieNode:
def __init__(self):
self.children = {}
self.end_of_word = False
def buildTrie(wordDict):
root = TrieNode()
for word in wordDict:
node = root
for char in word:
if char not in node.children:
node.children[char] = TrieNode()
node = node.children[char]
node.end_of_word = True
return root
def canBreak(s, root):
def search(start, node):
if start == len(s):
return True
if s[start] not in node.children:
return False
return search(start + 1, node.children[s[start]]) or (node.end_of_word and search(start, root))
return search(0, root)
word_list = ["apple", "pie", "banana"]
trie_root = buildTrie(word_list)
string_check = "applepie"
print(canBreak(string_check, trie_root))Output: True
The code defines a TrieNode class and a function to build a trie called buildTrie. The function canBreak performs a recursive search in the trie, using the function search. It returns True if and only if there is a way to segment the entire string using words from the trie.
Bonus One-Liner Method 5: Using Regular Expressions
A rapid and clever technique to address this issue is to exploit Python’s regular expression capabilities. By combining the list of words into a pattern string and using re.fullmatch, we can check if the entire string matches the pattern of word concatenations.
Here’s an example:
import re
def canBreak(s, wordDict):
pattern = '^' + '|'.join(map(re.escape, wordDict)) + '+$'
return bool(re.fullmatch(pattern, s))
word_list = ["apple", "pie", "banana"]
string_check = "applepie"
print(canBreak(string_check, word_list))Output: True
The canBreak function creates a regular expression pattern that matches any sequence of the words in the wordDict. The re.fullmatch function is used to verify if the entire string s conforms to the pattern.
Summary/Discussion
- Method 1: Recursive Approach with Memoization. This method is simple and straightforward, but its performance highly depends on the size of the input and can lead to a stack overflow for large inputs.
- Method 2: Dynamic Programming. This method offers a bottom-up solution that is efficient in terms of space and time complexity. However, it might require a higher understanding of dynamic programming concepts.
- Method 3: Breadth-First Search. BFS is a graph-based approach, which is useful for exploration of all possibilities but might become slow if the input string or word list is very large.
- Method 4: Using Trie Data Structure. Trie structure provides an excellent way to handle the problem with prefix searches being optimal, although it requires additional work to set up the trie structure.
- Bonus Method 5: Using Regular Expressions. Regex is a powerful one-liner solution that’s easy to implement but might not be as efficient for very large lists of words or long strings.