Problem
Decide whether text can be split into a sequence of words from words. Words may be reused. Dynamic programming over prefixes solves it efficiently.
Implement can_break(text, words).
Input
text: Lowercase letters only. words: The dictionary; each word has 1–50 lowercase letters.
Output
Return true if a split exists.
Constraints
0 ≤ length of text ≤ 5,000; 1 ≤ words ≤ 1,000
Examples
Input:
text = "applepenapple"
words = ["apple", "pen"]
Expected output:
expected = TrueInput:
std::string text = "applepenapple";
std::vector<std::string> words = {"apple", "pen"};
Expected output:
bool expected = true;Input:
string text = "applepenapple";
string[] words = new string[] {"apple", "pen"};
Expected output:
bool expected = true;Explanation
"apple" + "pen" + "apple", reusing "apple".
Input:
text = "catsandog"
words = ["cats", "dog", "sand", "and", "cat"]
Expected output:
expected = FalseInput:
std::string text = "catsandog";
std::vector<std::string> words = {"cats", "dog", "sand", "and", "cat"};
Expected output:
bool expected = false;Input:
string text = "catsandog";
string[] words = new string[] {"cats", "dog", "sand", "and", "cat"};
Expected output:
bool expected = false;Explanation
Every split leaves "og" or "ando" unmatched: false.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
A prefix is breakable if some earlier breakable point reaches it with a dictionary word.
Hint 2
Dynamic programming over positions works well when you test word lengths or dictionary matches efficiently.
Hint 3
Empty text is trivially breakable, but repeated word reuse should remain allowed throughout the scan.
Expected complexity
O(n^2) time, O(n) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.