What is the time complexity of Longest Word in Dictionary?

The optimal solution has a time complexity of O(NlogN).

What data structures are used to solve Longest Word in Dictionary?

This problem uses Array, Hash Table, String, Trie, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest Word in Dictionary?

Insight 1: Sorting the input words lexicographically allows us to prioritize smaller words and handle the lexicographical requirement in case of ties in length. Insight 2: Using a set to keep track of 'built' words efficiently allows us to check if a word's prefix exists in O(1) on average.

Which companies ask LeetCode 720. Longest Word in Dictionary in interviews?

This problem is frequently asked at Pinterest.

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Longest Word in Dictionary - LeetCode 720 Solution

Longest Word in Dictionary - Complete Solution Guide

Longest Word in Dictionary is LeetCode problem 720, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an array of strings words representing an English Dictionary, return the longest word in words that can be built one character at a time by other words in words . If there is more than one possible answer, return the longest word with the smallest lexicographical order. If there is no answer, return the empty string. Note that the word should be built from left to right with each additional character being added to the end of a previous word. Example 1: Input: words = ["w","wo","wor","worl

Detailed Explanation

The problem asks us to find the longest word in a given list of words (dictionary) that can be built one character at a time by other words also present in the list. The words are built sequentially, meaning each new character is appended to a valid existing word in the dictionary. If multiple words satisfy this condition and have the same length, we must return the lexicographically smallest one. If no such word exists, return an empty string.

Solution Approach

The solution first sorts the input `words` lexicographically. It then iterates through the sorted `words` list, maintaining a set called `built_words` which stores words that can be built by sequentially adding characters from other words in the list. For each `word`, it checks if its prefix (all characters except the last one) exists in `built_words`. If the prefix exists, the current `word` is added to `built_words` and its length is compared to the current `result`. If the current `word` is longer than the `result`, or if it's the same length but lexicographically smaller (handled by sorting), the `result` is updated.

Step-by-Step Algorithm

Step 1: Sort the input list of `words` lexicographically. This ensures the lexicographical tie-breaking rule is satisfied.
Step 2: Initialize an empty set called `built_words` and add an empty string to it. The empty string represents the base case for building words.
Step 3: Initialize an empty string called `result` to store the longest word found so far.
Step 4: Iterate through each `word` in the sorted `words` list.
Step 5: For each `word`, extract its prefix (the word without the last character).
Step 6: Check if the prefix exists in the `built_words` set.
Step 7: If the prefix exists in `built_words`, add the current `word` to the `built_words` set.
Step 8: If the length of the current `word` is greater than the length of the `result`, update `result` to be the current `word`.
Step 9: After iterating through all `words`, return the `result` string.

Key Insights

Insight 1: Sorting the input words lexicographically allows us to prioritize smaller words and handle the lexicographical requirement in case of ties in length.
Insight 2: Using a set to keep track of 'built' words efficiently allows us to check if a word's prefix exists in O(1) on average.
Insight 3: The empty string is implicitly 'built', serving as the base case for single-character words.

Complexity Analysis

Time Complexity: O(NlogN)

Space Complexity: O(N)

Topics

This problem involves: Array, Hash Table, String, Trie, Sorting.

Companies

Asked at: Pinterest.

Longest Word in Dictionary | LeetCode 720

Medium

Array Hash Table String Trie Sorting

Time: O(NlogN)Space: O(N)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

class Solution {
    public String longestWord(String[] words) {
        Arrays.sort(words);
        
        Set<String> builtWords = new HashSet<>();
        builtWords.add("");
        
        String result = "";
        
        for (String word : words) {
            if (builtWords.contains(word.substring(0, word.length() - 1))) {
                builtWords.add(word);
                if (word.length() > result.length()) {
                    result = word;
                }
            }
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 720: Longest Word in Dictionary - Detailed explanation and solution approaches.

Topics

Array Hash Table String Trie Sorting

Companies

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:720

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Sort the input list of `words` lexicographically. This ensures the lexicographical tie-breaking rule is satisfied.
2
Step 2: Initialize an empty set called `built_words` and add an empty string to it. The empty string represents the base case for building words.
3
Step 3: Initialize an empty string called `result` to store the longest word found so far.
4
Step 4: Iterate through each `word` in the sorted `words` list.
5
Step 5: For each `word`, extract its prefix (the word without the last character).
6
Step 6: Check if the prefix exists in the `built_words` set.
7
Step 7: If the prefix exists in `built_words`, add the current `word` to the `built_words` set.
8
Step 8: If the length of the current `word` is greater than the length of the `result`, update `result` to be the current `word`.
9
Step 9: After iterating through all `words`, return the `result` string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(NlogN)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is dominated by the `Arrays.sort()` method. The space complexity is determined by the maximum size of the `builtWords` set, which can contain all the words if they all form a valid chain.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Trie

A Trie (prefix tree) can be used to store all the words. Then, a depth-first search (DFS) or breadth-first search (BFS) can be performed on the Trie to find the longest word that can be built one character at a time.

Time: O(N*L) where N is the number of words and L is the average length of the words for building the Trie. The search would be O(N) in the worst case, but less if branches die off early.Space: O(N*L) since Trie can have at most as many nodes as total characters in all words.

Trade-offs: The Trie approach can be more efficient if the words share many common prefixes, as it avoids redundant prefix checks. However, it requires more complex implementation and can have a higher space overhead if the prefixes are not commonly shared.

✨

Key Insights

1Insight 1: Sorting the input words lexicographically allows us to prioritize smaller words and handle the lexicographical requirement in case of ties in length.
2Insight 2: Using a set to keep track of 'built' words efficiently allows us to check if a word's prefix exists in O(1) on average.
3Insight 3: The empty string is implicitly 'built', serving as the base case for single-character words.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The solution correctly handles this case by returning an empty string since the loop won't execute.
Edge case 2: The input array contains only a single word. The solution correctly handles this case if the single word is a single character, because it will build from the empty string. If the word is longer and no prefix is present, an empty string will be returned.
Edge case 3: The input array contains words that can be built in multiple ways, but only one way leads to the longest word. The sorting ensures that if multiple solutions exist, the lexicographically smallest one is chosen, and the `built_words` set ensures that the longest buildable word is selected.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the words lexicographically, leading to incorrect results when there are multiple words of the same maximum length. The prompt states that the smaller lexicographical word must be returned if there is a tie.

✗Common mistake 2: Not considering the empty string as a valid starting point for building words, which would cause single-character words to be missed.

✗Common mistake 3: Incorrectly implementing the prefix check, especially in languages like C where manual memory management is required for substring operations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Spell checkers. When suggesting corrections for misspelled words, spell checkers can use a similar approach to find valid words that can be built by adding or removing characters from the misspelled word. The 'dictionary' would be all allowed words.

🏦Real-world scenario 2: Autocompletion systems. Similar to spell checkers, autocompletion systems can use prefix-based searching to suggest words that can be built from the user's input, providing a list of possible completions as the user types.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Longest Word in Dictionary"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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