How do you solve Replace Words on LeetCode?

Replace Words (LeetCode #648) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * k) space complexity. Key insight: Using a Trie allows for efficient prefix searching.

What is the brute force solution for Replace Words?

The brute force approach for Replace Words is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each word in the sentence against all roots in the dictionary. If a word starts with any root, it replaces that word with the root. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Replace Words?

Replace Words uses the following concepts: Array, Hash Table, String, Trie. The recommended patterns to study are: Trie, Prefix Tree.

What are the interview tips for Replace Words?

Explain your thought process clearly while coding. Discuss edge cases and how your solution handles them. Be prepared to optimize your solution if asked.

What are common mistakes when solving Replace Words?

Not considering all roots when replacing a word. Failing to handle cases where no root matches.

#648

Replace Words

Medium

ArrayHash TableStringTrieTriePrefix Tree

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n * k)

💡

Intuition

Time O(n * m)Space O(n * k)

Using a Trie (prefix tree) allows us to efficiently search for roots that match the beginning of each word in the sentence. This reduces the number of comparisons needed and speeds up the process significantly.

⚙️

Algorithm

5 steps

1Step 1: Build a Trie from the dictionary of roots.
2Step 2: Split the sentence into individual words.
3Step 3: For each word, traverse the Trie to find the shortest root that matches the start of the word.
4Step 4: Replace the word with the found root if it exists.
5Step 5: Join the modified words back into a sentence.

solution.py38 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.is_end = False
5
6class Trie:
7    def __init__(self):
8        self.root = TrieNode()
9
10    def insert(self, word):
11        node = self.root
12        for char in word:
13            if char not in node.children:
14                node.children[char] = TrieNode()
15            node = node.children[char]
16        node.is_end = True
17
18    def search(self, word):
19        node = self.root
20        for char in word:
21            if char not in node.children:
22                return ''
23            node = node.children[char]
24            if node.is_end:
25                return word[:word.index(char)+1]
26        return ''
27
28
29def replaceWords(dictionary, sentence):
30    trie = Trie()
31    for root in dictionary:
32        trie.insert(root)
33    words = sentence.split()
34    for i in range(len(words)):
35        root = trie.search(words[i])
36        if root:
37            words[i] = root
38    return ' '.join(words)

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of words in the sentence and m is the average length of the words. The space complexity is O(n * k) where k is the average length of the roots stored in the Trie.

1Using a Trie allows for efficient prefix searching.
2Replacing words with roots improves clarity and reduces redundancy.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.