Solution: Word Break

Explore how to apply dynamic programming to solve the Word Break problem in C++. This lesson guides you through building a lookup table to check string segmentation against a dictionary, improving efficiency by avoiding redundant checks. By the end, you'll understand the algorithm's time and space complexity and be able to implement this common interview pattern effectively.

We'll cover the following...

Statement
Solution
- Naive approach
- Optimized approach using dynamic programming

Statement

Given a string, s, and a dictionary of strings, wordDict, check if s can be segmented into a space-separated sequence of one or more dictionary words. If yes, return TRUE; else, return FALSE.

Note: The same word in the dictionary may be used multiple times.

Constraints:

$1 \leq$ s.length $\leq 250$
$1 \leq$ wordDict.length $\leq 1000$
$1 \leq$ wordDict[i].length $\leq 20$
s and wordDict[i] consist of only lowercase English letters.
All the strings of wordDict are unique.

Solution

So far, you’ve probably brainstormed some approaches and have an idea of how to solve this problem. Let’s explore some of these approaches and figure out which one to follow ...