What is the time complexity of Word Break II?

The optimal solution has a time complexity of O(2^n).

What data structures are used to solve Word Break II?

This problem uses Array, Hash Table, String, Dynamic Programming, Backtracking, Trie, Memoization. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Word Break II?

Insight 1: Backtracking with memoization is essential to avoid redundant computations and handle the potentially large number of possible segmentations, making it computationally feasible. Insight 2: The problem can be naturally solved using recursion, exploring all possible prefixes of the string and checking if they are valid words.

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Word Break II - LeetCode 140 Solution

Q: What is the space complexity of Word Break II?

The space complexity is O(n^3).

Q: Which companies ask LeetCode 140. Word Break II in interviews?

This problem is frequently asked at Dropbox, Grammarly, Moveworks, Oracle, ServiceNow and 4 more companies.

Word Break II - Complete Solution Guide

Word Break II is LeetCode problem 140, 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 a string s and a dictionary of strings wordDict , add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order . Note that the same word in the dictionary may be reused multiple times in the segmentation. Example 1: Input: s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"] Output: ["cats and dog","cat sand dog"] Example 2: Input: s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple

Detailed Explanation

The problem asks us to take a string `s` and a dictionary of words `wordDict`. The goal is to break the string `s` into a sequence of words, where each word is present in the `wordDict`. We need to return all possible ways to break the string into valid sentences, represented as a list of strings. The same word in `wordDict` can be reused multiple times. If no such segmentation is possible, we return an empty list.

Solution Approach

The provided solution uses a backtracking algorithm with memoization to find all possible word breaks. It explores the string recursively, attempting to match prefixes with words in the dictionary. Memoization is used to store the results of subproblems, preventing redundant computations and significantly improving performance. The base case is when the starting index reaches the end of the string, in which case an empty string is added to the result. The recursive step involves iterating through all possible ending indices to create candidate words. If a candidate word is in the dictionary, the algorithm recursively calls itself on the remaining part of the string, and combines the current word with the results of the recursive call to build valid sentences.

Step-by-Step Algorithm

Step 1: Create a set `word_set` from the `wordDict` for efficient word lookups.
Step 2: Initialize a memoization dictionary `memo` to store results of already computed subproblems where key is the start index and value is the list of possible sentences.
Step 3: Define a recursive function `backtrack(start_index)` that takes the starting index of the substring as input.
Step 4: Inside `backtrack`, check if the result for `start_index` is already memoized. If so, return the memoized value.
Step 5: If `start_index` reaches the end of the string, return a list containing an empty string (representing the base case: no more characters to process).
Step 6: Iterate from `start_index` to the end of the string, creating potential words (substrings).
Step 7: For each potential word, check if it's present in `word_set`.
Step 8: If the potential word is valid, recursively call `backtrack(end_index + 1)` to find all possible sentences for the remaining substring.
Step 9: Combine the current word with each of the sentences returned by the recursive call, adding a space if the returned sentence is not empty. Add the combined sentences to the `results` list.
Step 10: Store the computed `results` in the `memo` for `start_index` and return the `results` list.

Key Insights

Insight 1: Backtracking with memoization is essential to avoid redundant computations and handle the potentially large number of possible segmentations, making it computationally feasible.
Insight 2: The problem can be naturally solved using recursion, exploring all possible prefixes of the string and checking if they are valid words.
Insight 3: Using a set data structure for the word dictionary allows for efficient lookups (O(1) on average) when checking if a substring is a valid word.

Complexity Analysis

Time Complexity: O(2^n)

Space Complexity: O(n^3)

Topics

This problem involves: Array, Hash Table, String, Dynamic Programming, Backtracking, Trie, Memoization.

Companies

Asked at: Dropbox, Grammarly, Moveworks, Oracle, ServiceNow, Snap, TikTok, Uber, X.

Word Break II | LeetCode 140

Hard

Array Hash Table String Dynamic Programming Backtracking Trie Memoization

Time: O(2^n)Space: O(n^3)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<String> wordBreak(String s, List<String> wordDict) {
        Set<String> wordSet = new HashSet<>(wordDict);
        Map<Integer, List<String>> memo = new HashMap<>();

        return backtrack(s, wordSet, 0, memo);
    }

    private List<String> backtrack(String s, Set<String> wordSet, int startIndex, Map<Integer, List<String>> memo) {
        if (memo.containsKey(startIndex)) {
            return memo.get(startIndex);
        }

        if (startIndex == s.length()) {
            return Arrays.asList("");
        }

        List<String> results = new ArrayList<>();
        for (int endIndex = startIndex; endIndex < s.length(); endIndex++) {
            String word = s.substring(startIndex, endIndex + 1);
            if (wordSet.contains(word)) {
                List<String> tails = backtrack(s, wordSet, endIndex + 1, memo);
                for (String tail : tails) {
                    results.add(word + (tail.isEmpty() ? "" : " " + tail));
                }
            }
        }

        memo.put(startIndex, results);
        return results;
    }
}

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Problem Info

Description

Problem 140: Word Break II - Detailed explanation and solution approaches.

Topics

Array Hash Table String Dynamic Programming Backtracking Trie Memoization

Companies

DropboxGrammarlyMoveworksOracleServiceNowSnapTikTokUber

Stats

Difficulty:Hard

Languages:4

Companies:9

Problem #:140

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

10 steps

1
Step 1: Create a set `word_set` from the `wordDict` for efficient word lookups.
2
Step 2: Initialize a memoization dictionary `memo` to store results of already computed subproblems where key is the start index and value is the list of possible sentences.
3
Step 3: Define a recursive function `backtrack(start_index)` that takes the starting index of the substring as input.
4
Step 4: Inside `backtrack`, check if the result for `start_index` is already memoized. If so, return the memoized value.
5
Step 5: If `start_index` reaches the end of the string, return a list containing an empty string (representing the base case: no more characters to process).
6
Step 6: Iterate from `start_index` to the end of the string, creating potential words (substrings).
7
Step 7: For each potential word, check if it's present in `word_set`.
8
Step 8: If the potential word is valid, recursively call `backtrack(end_index + 1)` to find all possible sentences for the remaining substring.
9
Step 9: Combine the current word with each of the sentences returned by the recursive call, adding a space if the returned sentence is not empty. Add the combined sentences to the `results` list.
10
Step 10: Store the computed `results` in the `memo` for `start_index` and return the `results` 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(2^n)

🐌 Exponential - Very slow! Often needs optimization.

💾

Space Complexity

O(n^3)

📝

Why?

The time complexity is exponential in the worst case because the backtracking algorithm explores all possible ways to segment the string. Memoization improves the practical performance but does not change the theoretical worst-case complexity. The space complexity is dominated by the memoization table, which stores intermediate results. In the worst case, the memo table can hold close to n results, and each result could be a string of size n, therefore the space complexity is O(n^3).

📖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

Dynamic Programming (Bottom-up)

A bottom-up dynamic programming approach can also solve this problem. It involves building a table `dp` where `dp[i]` stores all possible sentences for the substring `s[0...i-1]`. The table is built iteratively, starting from the beginning of the string.

Time: O(n^3) in the worst case. Where 'n' is the length of the string. This complexity arises from the three nested loops: iterating through the string, iterating through possible starting points for words ending at the current position, and copying strings.Space: O(n^3). The space complexity is dominated by the `dp` table, which stores at most O(n) possible sentences each of length at most O(n^2).

Trade-offs: The dynamic programming approach can sometimes be more efficient than backtracking for certain inputs, especially when there are many overlapping subproblems. However, it can also be less intuitive to implement and may require more memory. Backtracking with memoization is often preferred for its relative simplicity and good performance in many practical cases.

✨

Key Insights

1Insight 1: Backtracking with memoization is essential to avoid redundant computations and handle the potentially large number of possible segmentations, making it computationally feasible.
2Insight 2: The problem can be naturally solved using recursion, exploring all possible prefixes of the string and checking if they are valid words.
3Insight 3: Using a set data structure for the word dictionary allows for efficient lookups (O(1) on average) when checking if a substring is a valid word.

📚

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!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Empty input string `s`. The algorithm handles this gracefully as the base case of the recursion, immediately returning a list containing an empty string, if the `wordDict` contains empty string or returning empty list if the dictionary doesn't contain it.
Edge case 2: No possible segmentation. If the string `s` cannot be segmented into words from `wordDict`, the algorithm will explore all possibilities and eventually return an empty list.
Edge case 3: A very long string and a large dictionary. Memoization helps to mitigate the performance impact, but for extremely large inputs, the algorithm could still be slow or exceed memory limits.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization. Without memoization, the algorithm can become extremely slow due to redundant computations, leading to a Time Limit Exceeded (TLE) error.

✗Common mistake 2: Incorrectly handling the base case. Failing to properly handle the base case of the recursion (when the start index reaches the end of the string) can lead to incorrect results or infinite recursion.

✗Common mistake 3: Forgetting to handle cases where there is no valid segmentation. The algorithm must be able to correctly identify and return an empty list when no valid segmentation is possible.

✗Common mistake 4: Memory leaks (especially in C). Failing to free allocated memory for substrings and the `results` array, especially after making recursive calls, can lead to memory leaks. Ensuring all dynamically allocated memory is freed is crucial.

✅

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 processing and natural language processing. This problem is relevant to tasks like sentence parsing, text segmentation, and natural language understanding, where it's necessary to break down text into meaningful units.

🏦Real-world scenario 2: Code compilation. Similar concepts are used in compilers to parse and break down source code into tokens, which are then used for further analysis and code generation.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: October 19, 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 13

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

@AlgoMasterJul 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.

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.

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 "Word Break II"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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