What is the time complexity of Concatenated Words?

The optimal solution has a time complexity of O(N * L^3).

What is the space complexity of Concatenated Words?

The space complexity is O(N * L).

What data structures are used to solve Concatenated Words?

This problem uses Array, String, Dynamic Programming, Depth-First Search, Trie, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Concatenated Words?

Insight 1: Sorting the words by length helps to efficiently build up the Trie and search for concatenated words. Shorter words can be added to the Trie first, ensuring that when a longer word is checked, all possible component words are already in the Trie. Insight 2: Using Dynamic Programming (DP) to check if a word is concatenated is crucial for efficiency. The DP approach avoids redundant checks by storing intermediate results, indicating whether a prefix of the word can be formed by concatenating shorter words.

Which companies ask LeetCode 472. Concatenated Words in interviews?

This problem is frequently asked at eBay.

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Concatenated Words - LeetCode 472 Solution

Concatenated Words - Complete Solution Guide

Concatenated Words is LeetCode problem 472, a Hard 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 ( without duplicates ), return all the concatenated words in the given list of words . A concatenated word is defined as a string that is comprised entirely of at least two shorter words (not necessarily distinct) in the given array. Example 1: Input: words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"] Output: ["catsdogcats","dogcatsdog","ratcatdogcat"] Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "c

Detailed Explanation

The problem asks us to identify concatenated words from a given list of strings. A concatenated word is a word that can be formed by combining at least two shorter words from the same list. The words in the list are unique, and the concatenated words may be formed by using the same word multiple times.

Solution Approach

The solution combines sorting, a Trie data structure, and dynamic programming. First, the words are sorted by their length in ascending order. This ensures that when checking for concatenated words, all possible shorter words that could make up the concatenated word are already present in the Trie. A Trie is used to store the words seen so far. For each word, a dynamic programming approach determines if it can be constructed from shorter words present in the Trie. If it can, it's added to the result list. The word is then added to the Trie so other longer words can reuse it.

Step-by-Step Algorithm

Step 1: Sort the input array of strings `words` by length in ascending order.
Step 2: Initialize an empty Trie data structure to store the words.
Step 3: Initialize an empty list `result` to store the concatenated words.
Step 4: Iterate through each word in the sorted `words` array.
Step 5: For each word, use dynamic programming to check if it can be formed by concatenating shorter words already present in the Trie.
Step 6: If the dynamic programming check returns true (i.e., the word is a concatenated word), add the word to the `result` list.
Step 7: Add the current word to the Trie.
Step 8: After iterating through all words, return the `result` list.

Key Insights

Insight 1: Sorting the words by length helps to efficiently build up the Trie and search for concatenated words. Shorter words can be added to the Trie first, ensuring that when a longer word is checked, all possible component words are already in the Trie.
Insight 2: Using Dynamic Programming (DP) to check if a word is concatenated is crucial for efficiency. The DP approach avoids redundant checks by storing intermediate results, indicating whether a prefix of the word can be formed by concatenating shorter words.
Insight 3: Trie data structure is used to efficiently determine if a substring of a word is present in the list of words.
Insight 4: A word cannot be constructed by itself according to the definition in the question. Thus, when finding smaller words in Trie, a concatenated word must contains at least two smaller words.

Complexity Analysis

Time Complexity: O(N * L^3)

Space Complexity: O(N * L)

Topics

This problem involves: Array, String, Dynamic Programming, Depth-First Search, Trie, Sorting.

Companies

Asked at: eBay.

Concatenated Words | LeetCode 472

Hard

Array String Dynamic Programming Depth-First Search Trie Sorting

Time: O(N * L^3)Space: O(N * L)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<String> findAllConcatenatedWordsInADict(String[] words) {
        
        class TrieNode {
            Map<Character, TrieNode> children;
            boolean is_end_of_word;

            public TrieNode() {
                this.children = new HashMap<>();
                this.is_end_of_word = false;
            }
        }

        TrieNode root = new TrieNode();

        
        
        
        

        
        
        
        
        
        
        

        
        
        
        
        
        
        
        

        
        
        
        
        
        
        
        

        
        
        
        
        
        
        
        

        
        
        
        
        
        
        
        

        
        

        

        

        

        

        

        

        

        

        

        

        

        
        

        

        

        

        

        

        

        

        

        

        

        

        

        
        

        

        

        

        

        

        

        

        

        

        

        

        

        
        
        
        
        
        
        

        
        
        
        

        
        
        
        

        
        
        

        
        
        

        
        
        

        
        
        
        

        
        
        
        

        
        
        
        

        
        
        
        

        
        
        

        
        
        

        
        
        

        
        

        

        

        

        

        

        

        

        

        

        

        

        
        

        

        

        

        

        

        

        

        

        

        

        

        

        
        

        

        

        

        

        

        

        

        

        

        

        

        

        TrieNode finalRoot = root;
        
        
        
        
        
        
        

        
        
        
        

        
        
        
        

        
        
        

        
        
        

        
        
        

        
        
        
        

        
        
        
        

        
        
        
        

        
        
        
        

        
        
        

        
        
        

        
        
        

        
        
        

        
        

        

        

        

        

        

        

        

        

        

        

        

        
        

        

        

        

        

        

        

        

        

        

        

        

        

        
        

        

        

        

        

        

        

        

        

        

        

        

        

        TrieNode finalRoot1 = root;

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 472: Concatenated Words - Detailed explanation and solution approaches.

Topics

Array String Dynamic Programming Depth-First Search Trie Sorting

Companies

eBay

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:472

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

8 steps

1
Step 1: Sort the input array of strings `words` by length in ascending order.
2
Step 2: Initialize an empty Trie data structure to store the words.
3
Step 3: Initialize an empty list `result` to store the concatenated words.
4
Step 4: Iterate through each word in the sorted `words` array.
5
Step 5: For each word, use dynamic programming to check if it can be formed by concatenating shorter words already present in the Trie.
6
Step 6: If the dynamic programming check returns true (i.e., the word is a concatenated word), add the word to the `result` list.
7
Step 7: Add the current word to the Trie.
8
Step 8: After iterating through all words, return the `result` list.

💡

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(N * L^3)

💾

Space Complexity

O(N * L)

📝

Why?

The time complexity is determined by the nested loops and string operations within the `check` function (DP). The outer loop in `check` iterates up to 'n' (length of the word), and the inner loop also iterates up to 'n'. Each inner loop iteration checks for a substring in the Trie, taking O(L) time in the worst case, leading to O(L^3). The space complexity is influenced by the Trie's storage requirements, which are proportional to the total characters across all words in the worst-case.

📖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

Depth First Search (DFS) without Trie

Instead of using a Trie, one could use a set to store the words encountered so far and recursively search for prefixes in that set using DFS. This approach has a simpler implementation.

Time: O(N * L^2) in the average case, O(N * L^3) in the worst case (similar to the main solution, but without the Trie's efficient prefix lookup).Space: O(N) to store the set of words.

Trade-offs: This approach can be simpler to implement, but may not be as efficient as the Trie-based solution, especially for large datasets. It avoids Trie construction overhead, but sacrifices efficient prefix searching.

✨

Key Insights

1Insight 1: Sorting the words by length helps to efficiently build up the Trie and search for concatenated words. Shorter words can be added to the Trie first, ensuring that when a longer word is checked, all possible component words are already in the Trie.
2Insight 2: Using Dynamic Programming (DP) to check if a word is concatenated is crucial for efficiency. The DP approach avoids redundant checks by storing intermediate results, indicating whether a prefix of the word can be formed by concatenating shorter words.
3Insight 3: Trie data structure is used to efficiently determine if a substring of a word is present in the list of words.
4Insight 4: A word cannot be constructed by itself according to the definition in the question. Thus, when finding smaller words in Trie, a concatenated word must contains at least two smaller words.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

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 string. The provided solutions explicitly handle empty strings by skipping them, preventing errors in the Trie operations and ensuring correct results.
Edge case 2: A word that can be formed by concatenating itself multiple times (e.g., 'abab' formed by 'ab' + 'ab'). The dynamic programming approach correctly handles this because it considers all possible concatenations of shorter words.
Edge case 3: Input with no concatenated words. The algorithm correctly returns an empty list in this scenario.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the words by length. Failing to sort results in incorrect identification of concatenated words because longer words might be checked before their component shorter words are added to the Trie or set.

✗Common mistake 2: Not handling the empty string case. Empty strings can cause errors if not explicitly checked for and skipped.

✗Common mistake 3: Incorrect DP implementation. The dynamic programming approach requires careful initialization and iteration to correctly identify if a word can be formed by concatenation. Off-by-one errors or incorrect conditions can lead to wrong results.

✗Common mistake 4: Using the word itself when checking for concatenated word status. Each word should be constructed by at least two smaller words.

✅

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: Text segmentation. Identifying concatenated words is analogous to segmenting a text string into meaningful words, which is useful in natural language processing for tasks like tokenization and parsing.

🏦Real-world scenario 2: Code analysis. In programming languages, identifying concatenated identifiers or strings can be useful for code analysis and refactoring tools.

🎮Real-world scenario 3: Data compression. Identifying patterns in data that can be represented as concatenations of smaller units can be useful for data compression algorithms.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

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: September 21, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

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.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Concatenated Words"?

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

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