README.md

139. 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.

Solutions (Python)

1. Dynamic Programming

classSolution: defwordBreak(self, s: str, wordDict: List[str]) ->bool: dp= [True] + [False] *len(s) foriinrange(1, len(s) +1): forjinrange(i): ifdp[j] ands[j:i] inwordDict: dp[i] =Truebreakreturndp[-1]

Solutions (Ruby)

1. Dynamic Programming

# @param {String} s# @param {String[]} word_dict# @return {Boolean}defword_break(s,word_dict)dp=[true] + [false] * s.size(1..s.size).eachdo |i| (0...i).eachdo |j| ifdp[j] && word_dict.include?(s[j...i])dp[i]=truebreakendendenddp[s.size]end