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.
Input: s = "leetcode", wordDict = ["leet","code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code".
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.
Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"] Output: false
1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s
andwordDict[i]
consist of only lowercase English letters.- All the strings of
wordDict
are unique.
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
dp = [True] + [False] * len(s)
for i in range(1, len(s) + 1):
for j in range(i):
if dp[j] and s[j:i] in wordDict:
dp[i] = True
break
return dp[-1]
# @param {String} s
# @param {String[]} word_dict
# @return {Boolean}
def word_break(s, word_dict)
dp = [true] + [false] * s.size
(1..s.size).each do |i|
(0...i).each do |j|
if dp[j] && word_dict.include?(s[j...i])
dp[i] = true
break
end
end
end
dp[s.size]
end