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Word Break

Decide whether text can be split into dictionary words.

Problème

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.

Contraintes

0 ≤ length of text ≤ 5,000; 1 ≤ words ≤ 1,000

Exemples

Exemple 1
Entrée :
text = "applepenapple"
words = ["apple", "pen"]

Sortie attendue :
expected = True
Entrée :
std::string text = "applepenapple";
std::vector<std::string> words = {"apple", "pen"};

Sortie attendue :
bool expected = true;
Entrée :
string text = "applepenapple";
string[] words = new string[] {"apple", "pen"};

Sortie attendue :
bool expected = true;

Explication

"apple" + "pen" + "apple", reusing "apple".

Exemple 2
Entrée :
text = "catsandog"
words = ["cats", "dog", "sand", "and", "cat"]

Sortie attendue :
expected = False
Entrée :
std::string text = "catsandog";
std::vector<std::string> words = {"cats", "dog", "sand", "and", "cat"};

Sortie attendue :
bool expected = false;
Entrée :
string text = "catsandog";
string[] words = new string[] {"cats", "dog", "sand", "and", "cat"};

Sortie attendue :
bool expected = false;

Explication

Every split leaves "og" or "ando" unmatched: false.

Références

Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.

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