5 Best Ways to Check if a Target Word can be Spelled Out by a List of Words in Python

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πŸ’‘ Problem Formulation: We often encounter situations where we need to determine if a given target word can be constructed from a collection of available words. For example, given the list of words ['bat', 'ball', 'to'] and the target word 'bat', we want to check if ‘bat’ can be spelled out using the list, which in this case is true.

Method 1: Using Recursion

This method leverages the power of recursion to break down the problem into smaller instances, checking at each step if the target can be composed from the prefix of a word in the list and recursively checking the remainder of the target.

Here’s an example:

def can_spell(target, words):
    if target == '':
        return True
    for word in words:
        if target.startswith(word):
            remainder = target[len(word):]
            if can_spell(remainder, words):
                return True
    return False

words_list = ['bat', 'ball', 'to']
target_word = 'batt'
print(can_spell(target_word, words_list))

Output of this code snippet: False

The function can_spell first checks if the target is an empty string, indicating all words have been used to successfully spell out the target. It then iterates over the words list, checking if each word can serve as a prefix for the target. If found, the function calls itself recursively with the remainder of the target after removing the prefix.

Method 2: Using Dynamic Programming

Dynamic programming optimizes the recursive approach by storing intermediate results, which avoids re-computation and reduces time complexity, especially for larger word lists and targets.

Here’s an example:

def can_spell_dp(target, words):
    dp = [False] * (len(target) + 1)
    dp[0] = True
    for i in range(1, len(target) + 1):
        for word in words:
            if dp[i - len(word)] and target.startswith(word, i - len(word)):
                dp[i] = True
    return dp[len(target)]

words_list = ['bat', 'ball', 'to']
target_word = 'batto'
print(can_spell_dp(target_word, words_list))

Output of this code snippet: True

The function can_spell_dp creates a boolean array dp where each index represents if the target up to that point can be constructed. It iterates through the target and words list, updating the dp array where matches are found. The final value at dp[len(target)] gives the answer.

Method 3: Using Iterative Approach

This method uses an iterative approach that simulates the recursive stack to check if the target word can be built from the provided list of words, iterating the process from start to end of the target.

Here’s an example:

def can_spell_iterative(target, words):
    stack = [target]
    while stack:
        current_target = stack.pop()
        if current_target == '':
            return True
        for word in words:
            if current_target.startswith(word):
    return False

words_list = ['ball', 'ball', 'to', 'bat']
target_word = 'ballbat'
print(can_spell_iterative(target_word, words_list))

Output of this code snippet: True

The can_spell_iterative function uses a stack to simulate recursion. When a word in the list matches the start of the current target, the remainder of the target after removing the matching word is added back into the stack. The function returns true if it finds a way to reduce the entire target to an empty string.

Method 4: Using Trie Data Structure

Utilizing a trie (prefix tree) for storing words can optimize searching for prefixes and provide an efficient way to resolve the target word from the list.

Here’s an example:

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end_of_word = False

def insert_to_trie(root, word):
    node = 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 can_spell_with_trie(target, root):
    node = root
    for char in target:
        if char not in node.children:
            return False
        node = node.children[char]
    return node.is_end_of_word

# Build Trie with word list
trie_root = TrieNode()
for word in ['bat', 'ball', 'to']:
    insert_to_trie(trie_root, word)

print(can_spell_with_trie('ball', trie_root))

Output of this code snippet: True

The TrieNode class represents each node in the trie. The function insert_to_trie is used to insert words into the trie. can_spell_with_trie searches each letter of the target in the trie, returning whether the last letter is marked as an end of a word, thus determining if the target can be spelled out.

Bonus One-Liner Method 5: List Comprehension and All

A Pythonic one-liner approach utilizing list comprehension and the all function checks if all characters of the target are included in the word list.

Here’s an example:

words_list = ['bat', 'ball', 'to']
target_word = 'tab'
can_spell = all(target_word.count(char) <= ''.join(words_list).count(char) for char in set(target_word))

Output of this code snippet: True

This one-liner checks if the count of each unique character in the target word is less than or equal to the count of that character in the concatenated string of words. This approach assumes that the target word is case-sensitive and does not check for the order of characters.


  • Method 1: Recursion. Straightforward conceptually but can be inefficient for large inputs due to redundancy in computations.
  • Method 2: Dynamic Programming. More efficient, especially for larger inputs, by remembering past computations and avoiding recursion.
  • Method 3: Iterative Approach. It simulates recursion without the overhead but can become complex with extensive control structures.
  • Method 4: Trie Data Structure. Highly efficient for large datasets, especially when character lookups are frequent, though requires additional space for the trie.
  • Bonus Method 5: One-Liner List Comprehension. Quick and elegant Pythonic method but may not account for the sequence of characters in the target word.