💡 Problem Formulation: How can one determine the number of possible combinations to form a specific target string using a collection of sub-strings provided in a dictionary? For instance, if the target string is “coding” and the dictionary contains [“cod”, “ing”, “code”, “in”, “g”], we aim to find all unique ways this string can be constructed from the dictionary entries. The desired output is the total number of such unique combinations.
Method 1: Recursive Approach
This method entails a recursive function that goes through each dictionary word and sees if it is a prefix of the target string. If it is, the function is called recursively with the remainder of the string, accumulating the number of ways the target string can be formed. It’s a straightforward, brute-force technique, though not the most efficient.
Here’s an example:
def count_ways(target, words):
if target == '':
return 1
total_ways = 0
for word in words:
if target.startswith(word):
suffix = target[len(word):]
total_ways += count_ways(suffix, words)
return total_ways
dictionary = ["cod", "ing", "code", "in", "g"]
print(count_ways("coding", dictionary))Output: 2
The code snippet defines a function count_ways that recursively computes the number of ways to form the target string by prefix-matching dictionary words. The result of 2 indicates that there are two ways to form “coding” using the dictionary provided.
Method 2: Dynamic Programming (Top-Down)
Using dynamic programming, we can store intermediary results to prevent redundant computations, specifically with a top-down memoization approach. This method significantly reduces the time complexity, especially for larger strings and dictionaries, by caching the results of subproblems.
Here’s an example:
def count_ways(target, words, memo=None):
if memo is None:
memo = {}
if target in memo:
return memo[target]
if target == '':
return 1
total_ways = 0
for word in words:
if target.startswith(word):
suffix = target[len(word):]
total_ways += count_ways(suffix, words, memo)
memo[target] = total_ways
return total_ways
dictionary = ["cod", "ing", "code", "in", "g"]
print(count_ways("coding", dictionary))Output: 2
The count_ways function now includes a memo dictionary that caches the results of previously computed subproblems. This modification significantly optimizes the performance by avoiding redundant calculations.
Method 3: Dynamic Programming (Bottom-Up)
Another dynamic programming strategy known as bottom-up iteration avoids recursion altogether. This method iterates through the target string, building up the number of ways to form increasingly larger prefixes of the target string.
Here’s an example:
def count_ways(target, words):
ways = [0] * (len(target) + 1)
ways[0] = 1 # Empty string can always be formed in one way
for i in range(1, len(target) + 1):
for word in words:
if target[:i].endswith(word):
ways[i] += ways[i - len(word)]
return ways[len(target)]
dictionary = ["cod", "ing", "code", "in", "g"]
print(count_ways("coding", dictionary))Output: 2
The code initializes an array ways to track the number of formations for all prefixes of the string. The function then iteratively increases the count for each prefix that can be further constructed by adding a word from the dictionary.
Method 4: Using Trie
Incorporating a trie (or prefix tree) as a data structure can efficiently represent the dictionary, especially when dealing with a large number of dictionary words. The approach allows for quick checks of word prefixes while traversing the target string.
Here’s an example:
class TrieNode:
def __init__(self):
self.children = {}
self.end_of_word = False
def insert(root, word):
node = root
for letter in word:
if letter not in node.children:
node.children[letter] = TrieNode()
node = node.children[letter]
node.end_of_word = True
def count_ways(target, root, dp, index=0):
if index == len(target):
return 1
if dp[index] != -1:
return dp[index]
node = root
total_ways = 0
for i in range(index, len(target)):
if target[i] not in node.children:
break
node = node.children[target[i]]
if node.end_of_word:
total_ways += count_ways(target, root, dp, i+1)
dp[index] = total_ways
return total_ways
dictionary = ["cod", "ing", "code", "in", "g"]
root = TrieNode()
for word in dictionary:
insert(root, word)
dp = [-1] * (len("coding") + 1)
print(count_ways("coding", root, dp))Output: 2
In this snippet, a TrieNode class and insertion method are used to build a trie from the dictionary. The count_ways function then traverses the trie and target string simultaneously to count the number of ways the target can be formed. Dynamic programming is also used here to store intermediate results.
Bonus One-Liner Method 5: Using Built-In Combinations (For Limited Cases)
A one-liner approach using built-in Python functions like itertools.combinations is possible for small strings and dictionaries. This method is limited and more of a theoretical interest, as it does not scale to larger problems.
Here’s an example:
import itertools dictionary = ["cod", "ing", "code", "in", "g"] target = "coding" # This is only for showcasing and is not a practical solution for large cases. total_ways = sum(1 for combo in itertools.permutations(dictionary, len(target)) if ''.join(combo) == target) print(total_ways)
Output: 0
This line attempts to use permutations to find the total number of ways to form the target string. It will give a correct count in some trivial cases but fails for actual string combination problems due to the nature of permutations vs combinations.
Summary/Discussion
- Method 1: Recursive Approach. Simple to understand and implement; however, it has poor performance on long strings due to exponential time complexity and redundancy in calculations.
- Method 2: Dynamic Programming (Top-Down). Memoization improves efficiency over the recursive approach, suitable for medium-sized problems, still has overhead from recursion.
- Method 3: Dynamic Programming (Bottom-Up). Iterative solution that is typically more space and time efficient than top-down recursion and is well-suited for larger problems.
- Method 4: Using Trie. Excellent for a large set of dictionary words; provides an optimized way to traverse and check for prefixes, complemented with dynamic programming for best performance.
- Method 5: Built-In Combinations (One-Liner). A theoretical and limited approach, not practical for the problem but interesting to examine simple cases and understand combinatorial principles.