What is the time complexity of Word Break?

The optimal solution has a time complexity of O(n*m).

What is the space complexity of Word Break?

The space complexity is O(n+k).

What data structures are used to solve Word Break?

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

What are the key insights for solving Word Break?

Insight 1: Dynamic programming is effective for this problem because we can break down the problem into smaller subproblems. dp[i] represents whether the substring s[0...i-1] can be segmented into words from the dictionary. Insight 2: Using a set to store the dictionary words allows for efficient lookups (O(1) on average).

Which companies ask LeetCode 139. Word Break in interviews?

This problem is frequently asked at Block, Cleartrip, Coupang, CureFit, Flipkart and 20 more companies.

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Word Break - LeetCode 139 Solution

Word Break - Complete Solution Guide

Word Break is LeetCode problem 139, 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 a string s and a dictionary of strings wordDict , return true if s can be segmented into a space-separated sequence of one or more dictionary words. Note that the same word in the dictionary may be reused multiple times in the segmentation. Example 1: Input: s = "leetcode", wordDict = ["leet","code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code". Example 2: Input: s = "applepenapple", wordDict = ["apple","pen"] Output: true Explanation: Return tru

Detailed Explanation

The Word Break problem asks whether a given string `s` can be segmented into a sequence of one or more words from a given dictionary `wordDict`. The same word in the dictionary can be reused multiple times. The input consists of a string `s` and a list of strings `wordDict`. The output is a boolean value: `true` if `s` can be segmented, and `false` otherwise. For example, if `s` is "leetcode" and `wordDict` is ["leet", "code"], the answer is `true` because "leetcode" can be segmented as "leet code". The constraints include the length of `s` being between 1 and 300, the length of `wordDict` being between 1 and 1000, and the length of each word in `wordDict` being between 1 and 20. All strings consist of lowercase English letters, and the words in `wordDict` are unique.

Solution Approach

The solution employs dynamic programming to solve this problem. A boolean array `dp` of size `n+1` is used, where `dp[i]` is `true` if the substring `s[0...i-1]` can be segmented into words from the `wordDict`, and `false` otherwise. The base case is `dp[0] = true`, indicating that an empty string can be segmented. The algorithm iterates through the string `s`, and for each index `i`, it iterates from `j = max(0, i - max_len)` to `i-1`, where `max_len` is the maximum length of any word in `wordDict`. If `dp[j]` is `true` (meaning `s[0...j-1]` can be segmented) and the substring `s[j...i-1]` is in the `wordDict`, then `dp[i]` is set to `true`, indicating that `s[0...i-1]` can be segmented. The outer loop continues until the end of the string, and the final result is `dp[n]`. This approach correctly determines if the entire string `s` can be segmented into words from `wordDict`.

Step-by-Step Algorithm

Step 1: Create a set `word_set` from the given `wordDict` for fast lookup.
Step 2: Initialize a boolean array `dp` of size `n+1` with all values set to `false`, where `n` is the length of the string `s`. Set `dp[0] = true`.
Step 3: Find the maximum length `max_len` of any word in the `wordDict`. This optimization helps limit the inner loop's iterations.
Step 4: Iterate through the string `s` from index `i = 1` to `n`.
Step 5: For each index `i`, iterate from `j = max(0, i - max_len)` to `i - 1`.
Step 6: If `dp[j]` is `true` (meaning the substring `s[0...j-1]` can be segmented) and the substring `s[j:i]` is present in the `word_set`, then set `dp[i] = true` and break the inner loop.
Step 7: After the outer loop finishes, return `dp[n]`, which indicates whether the entire string `s` can be segmented.

Key Insights

Insight 1: Dynamic programming is effective for this problem because we can break down the problem into smaller subproblems. dp[i] represents whether the substring s[0...i-1] can be segmented into words from the dictionary.
Insight 2: Using a set to store the dictionary words allows for efficient lookups (O(1) on average).
Insight 3: Optimizing the inner loop by only checking substrings of length up to the maximum length of the words in the dictionary improves performance.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n+k)

Topics

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

Companies

Asked at: Block, Cleartrip, Coupang, CureFit, Flipkart, Grammarly, IBM, Intuit, MongoDB, Netflix, Netskope, Nordstrom, Nutanix, Otter.ai, Palo Alto Networks, Pocket Gems, SAP, Salesforce, Snowflake, TikTok, Uber, Walmart Labs, Yahoo, Zoho, eBay.

Word Break | LeetCode 139

Medium

Array Hash Table String Dynamic Programming Trie Memoization

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

Solution Code

Available in:

import java.util.List;
import java.util.Set;
import java.util.HashSet;

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> wordSet = new HashSet<>(wordDict);
        int n = s.length();
        boolean[] dp = new boolean[n + 1];
        dp[0] = true;

        int maxLen = 0;
        if (!wordSet.isEmpty()) {
            for (String w : wordSet) {
                maxLen = Math.max(maxLen, w.length());
            }
        }

        for (int i = 1; i <= n; i++) {
            for (int j = Math.max(0, i - maxLen); j < i; j++) {
                if (dp[j] && wordSet.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }

        return dp[n];
    }
}

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Problem Info

Description

Problem 139: Word Break - Detailed explanation and solution approaches.

Topics

Array Hash Table String Dynamic Programming Trie Memoization

Companies

BlockCleartripCoupangCureFitFlipkartGrammarlyIBMIntuit

Stats

Difficulty:Medium

Languages:4

Companies:25

Problem #:139

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

7 steps

1
Step 1: Create a set `word_set` from the given `wordDict` for fast lookup.
2
Step 2: Initialize a boolean array `dp` of size `n+1` with all values set to `false`, where `n` is the length of the string `s`. Set `dp[0] = true`.
3
Step 3: Find the maximum length `max_len` of any word in the `wordDict`. This optimization helps limit the inner loop's iterations.
4
Step 4: Iterate through the string `s` from index `i = 1` to `n`.
5
Step 5: For each index `i`, iterate from `j = max(0, i - max_len)` to `i - 1`.
6
Step 6: If `dp[j]` is `true` (meaning the substring `s[0...j-1]` can be segmented) and the substring `s[j:i]` is present in the `word_set`, then set `dp[i] = true` and break the inner loop.
7
Step 7: After the outer loop finishes, return `dp[n]`, which indicates whether the entire string `s` can be segmented.

💡

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*m)

💾

Space Complexity

O(n+k)

📝

Why?

The time complexity is determined by the nested loops. The outer loop runs `n` times, and the inner loop runs at most `m` times where `m` is the `max_len`. The space complexity is determined by the `dp` array of size `n+1` and the hash set `word_set` which at worst holds every string in `wordDict`.

📖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

Breadth-First Search (BFS)

Treat the string `s` as a graph, where each index is a node. An edge exists between two nodes `i` and `j` if the substring `s[i:j]` is in the dictionary. Perform a BFS starting from index 0 to reach index `n`.

Time: O(n^2) in the worst case (similar to the DP approach without the max length optimization) because we explore all possible substrings.Space: O(n), for the queue and visited set.

Trade-offs: BFS can be more intuitive for some, but the dynamic programming approach is generally more efficient, especially with the `max_len` optimization. BFS might have issues with very large dictionary or long strings due to memory allocation.

Recursion with Memoization

A recursive approach can be used where the function checks if the string can be segmented starting from a given index. Memoization stores the results of previously computed subproblems to avoid redundant calculations.

Time: O(n*m), where n is the length of s and m is the max length of a word in wordDict, the same as the DP solution.Space: O(n), for the memoization table and call stack.

Trade-offs: Recursion with memoization can be more readable than dynamic programming for some, but it can lead to stack overflow errors for very long strings.

✨

Key Insights

1Insight 1: Dynamic programming is effective for this problem because we can break down the problem into smaller subproblems. dp[i] represents whether the substring s[0...i-1] can be segmented into words from the dictionary.
2Insight 2: Using a set to store the dictionary words allows for efficient lookups (O(1) on average).
3Insight 3: Optimizing the inner loop by only checking substrings of length up to the maximum length of the words in the dictionary improves performance.

📚

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 string `s` and empty `wordDict`. The code correctly handles this case because `dp[0]` is initialized to `true`, and the outer loop doesn't execute, returning `dp[0]` which is `true` (although in the original problem description the constraint 1 <= s.length so s can't be empty).
Edge case 2: A long string `s` that cannot be segmented. The code will iterate through all possible substrings and correctly determine that `dp[n]` remains `false`.
Edge case 3: `wordDict` contains only very long words compared to the string `s`. The `max_len` optimization significantly reduces computations, as inner loop only runs on relevant lengths.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a set for the dictionary words. This leads to O(k) lookups in the inner loop, increasing the overall time complexity to O(n*m*k).

✗Common mistake 2: Not handling the base case `dp[0] = true` correctly. This will lead to incorrect results because the segmentation cannot start from an empty string.

✗Common mistake 3: Failing to optimize the inner loop with the `max_len` variable. Without this optimization, the inner loop iterates through all possible starting indices, regardless of the dictionary's contents, increasing the time complexity.

✗Common mistake 4: Memory management issues in C. Forgetting to free the allocated memory for the `sub` string can lead to memory leaks.

✅

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, where segmenting sentences into words or phrases is a common task. This algorithm can be used to check if a sentence can be formed from a known vocabulary.

🏦Real-world scenario 2: Code compilation, where breaking down a source code string into tokens (keywords, identifiers, operators) is a fundamental step. A similar approach could be used to check if the source code is syntactically valid based on a set of predefined tokens.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: October 18, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Word Break"?

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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