How do you solve Longest Word in Dictionary on LeetCode?

Longest Word in Dictionary (LeetCode #720) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting the words helps in checking prefixes in lexicographical order.

What is the time complexity of Longest Word in Dictionary?

The optimal solution for Longest Word in Dictionary runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the words in a set.

What algorithm or data structure is used to solve Longest Word in Dictionary?

Longest Word in Dictionary uses the following concepts: Array, Hash Table, String, Trie, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Word in Dictionary?

Always consider edge cases, such as single-character words. Explain your thought process clearly while coding. Don't forget to test your solution with multiple inputs.

What are common mistakes when solving Longest Word in Dictionary?

Not checking all prefixes correctly. Forgetting to handle the lexicographical order when lengths are the same.

Longest Word in Dictionary — LeetCode #720 (Medium)

Tags: Array, Hash Table, String, Trie, Sorting

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

In this approach, we will check each word in the list to see if all of its prefixes exist in the list. This is a straightforward method but can be inefficient as it requires multiple checks for each word.

The time complexity is O(n²) because for each word (n), we check all its prefixes (up to n). The space complexity is O(1) since we only use a few variables to store the longest word.

Step 1: Initialize a variable to store the longest valid word.
Step 2: For each word in the list, check if all prefixes of that word exist in the list.
Step 3: If a word is valid and longer than the current longest word, update the longest word.
Step 4: If two words are of the same length, keep the lexicographically smaller one.

1. Start with words = ["w", "wo", "wor", "worl", "world"]. 2. Initialize longest = "". 3. Check 'w': valid (longest = 'w'). 4. Check 'wo': valid (longest = 'wo'). 5. Check 'wor': valid (longest = 'wor'). 6. Check 'worl': valid (longest = 'worl'). 7. Check 'world': valid (longest = 'world'). Final longest = 'world'.

Optimal Solution approach

Time complexity: O(n log n). Space complexity: O(n).

This approach uses a set to store the words for quick lookup. By sorting the words first, we ensure that we can build longer words in lexicographical order efficiently.

The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the words in a set.

Step 1: Sort the words array lexicographically.
Step 2: Initialize a set to store the words and a variable for the longest word.
Step 3: Iterate through the sorted words and check if the current word can be formed by its prefixes in the set.
Step 4: If valid, update the longest word.
Step 5: Add the current word to the set.

1. Start with words = ["w", "wo", "wor", "worl", "world"]. 2. Sort words: ["w", "wo", "wor", "worl", "world"]. 3. Initialize longest = "" and wordSet = {}. 4. Check 'w': valid (longest = 'w'). Add 'w' to set. 5. Check 'wo': valid (longest = 'wo'). Add 'wo' to set. 6. Check 'wor': valid (longest = 'wor'). Add 'wor' to set. 7. Check 'worl': valid (longest = 'worl'). Add 'worl' to set. 8. Check 'world': valid (longest = 'world'). Add 'world' to set. Final longest = 'world'.

Key Insights

Sorting the words helps in checking prefixes in lexicographical order.
Using a set allows for O(1) average time complexity for lookups.

Common Mistakes

Not checking all prefixes correctly.
Forgetting to handle the lexicographical order when lengths are the same.

Interview Tips

Always consider edge cases, such as single-character words.
Explain your thought process clearly while coding.
Don't forget to test your solution with multiple inputs.

#720

Longest Word in Dictionary

Medium

Array↗Hash Table↗String↗Trie↗Sorting↗Hash Map↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

This approach uses a set to store the words for quick lookup. By sorting the words first, we ensure that we can build longer words in lexicographical order efficiently.

⚙️

Algorithm

5 steps

1Step 1: Sort the words array lexicographically.
2Step 2: Initialize a set to store the words and a variable for the longest word.
3Step 3: Iterate through the sorted words and check if the current word can be formed by its prefixes in the set.
4Step 4: If valid, update the longest word.
5Step 5: Add the current word to the set.

solution.py9 lines

1def longestWord(words):
2    words.sort()
3    word_set = set()
4    longest = ''
5    for word in words:
6        if len(word) == 1 or word[:-1] in word_set:
7            longest = word if len(word) > len(longest) else longest
8            word_set.add(word)
9    return longest

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the words in a set.

1Sorting the words helps in checking prefixes in lexicographical order.
2Using a set allows for O(1) average time complexity for lookups.

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