Word Break
Detailed guide and Python implementation for the 'Word Break' problem.
1. Concept Overview
The 'Word Break' problem is a key challenge in the 1D DP section.
This implementation focuses on medium-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Word Break.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Word Break carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
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.
Write a function wordBreak(s: str, wordDict: List[str]) -> bool.
- •1 <= len(s) <= 300
- •1 <= len(wordDict) <= 1000
- •1 <= len(wordDict[i]) <= 20
- •s and wordDict[i] consist of only lowercase English letters
- •All the strings of wordDict are unique
Examples
s = "leetcode", wordDict = ["leet","code"]
True
Return True because "leetcode" can be segmented as "leet code".
s = "applepenapple", wordDict = ["apple","pen"]
True
Return True because "applepenapple" can be segmented as "apple pen apple".
s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
False
No valid segmentation exists.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def word_break_opt(s, wordDict):
dp = [False] * (len(s) + 1)
dp[len(s)] = True
for i in range(len(s) - 1, -1, -1):
for w in wordDict:
if (i + len(w)) <= len(s) and s[i : i + len(w)] == w:
dp[i] = dp[i + len(w)]
if dp[i]: break
return dp[0]Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def word_break_brute(s, wordDict):
def solve(idx):
if idx == len(s): return True
for w in wordDict:
if s[idx:].startswith(w) and solve(idx + len(w)):
return True
return False
return solve(0)Algorithm Pattern Checklist
When dealing with 1D DP data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
Related Questions
Recommended Python Resources
Expand your knowledge with related interactive tutorials, cheat sheets, and code comparisons.
Python Loops
Learn how to use Python loops to iterate over data. Master for loops, while loops, break, continue, and loop best practices with interactive examples.
How to Sort a List in Python
Learn how to sort a list in Python using the sort() method and the sorted() function. Discover custom key sorting and reverse order examples.
Python String Methods
A complete reference guide for Python string manipulation. Master formatting, searching, splitting, replacing, and checking string properties.
Python vs JavaScript: Which Programming Language is Best?
A comprehensive comparison between Python and JavaScript. Explore syntax differences, performance, use cases (backend vs frontend), and coding examples.