Word Break

Problem

https://leetcode.com/problems/word-break/

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 true because "applepenapple" can be segmented as "apple pen apple".
Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
Output: false

Constraints:

• 1 <= s.length <= 300

• 1 <= wordDict.length <= 1000

• 1 <= wordDict[i].length <= 20

• s and wordDict[i] consist of only lowercase English letters.

• All the strings of wordDict are unique.

Pattern

Array, Hash Table, String, Dynamic Programming, Trie

Approaches

Memoization

Code

def wordBreak(s: str, wordDict: list[str]) -> bool:
    wordDict = set(wordDict)

    memo = {len(s): True}

    def word_break(i):
        if i in memo:
            return memo[i]

        result = False
        for j in range(i + 1, len(s) + 1):
            if s[i:j] in wordDict:
                result = result or word_break(j)

        memo[i] = result
        return result

    return word_break(0)

Test

>>> from word_break__memoization import wordBreak
>>> wordBreak("leetcode", ["leet", "code"])
True
>>> wordBreak("applepenapple", ["apple", "pen"])
True
>>> wordBreak("catsandog", ["cats", "dog", "sand", "and", "cat"])
False

word_break__memoization.wordBreak(s: str, wordDict: list[str]) → bool