LeetCode
LeetCode 140 Word Break II - Hard
140. Word Break II -- Hard
140. Word Break II — Hard
Problem
- Word Break II -- Hard
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.
Note:
The same word in the dictionary may be reused multiple times in the segmentation. You may assume the dictionary does not contain duplicate words.
Example 1: Input: s = "catsanddog" wordDict = ["cat", "cats", "and", "sand", "dog"] Output: [ "cats and dog", "cat sand dog" ]
Example 2: Input: s = "pineapplepenapple" wordDict = ["apple", "pen", "applepen", "pine", "pineapple"] Output: [ "pine apple pen apple", "pineapple pen apple", "pine applepen apple" ] Explanation: Note that you are allowed to reuse a dictionary word.
Example 3: Input: s = "catsandog" wordDict = ["cats", "dog", "sand", "and", "cat"] Output: []
Solution
class Solution:
def __init__(self):
self.res = []
def dfs(self, s, path, dp, ind, word_dict):
if not dp[ind + len(s)]:
return
if dp[ind+len(s)]:
if not s:
self.res.append(path.strip())
return
for i in range(1, len(s)+1):
if s[:i] in word_dict:
self.dfs(s[i:], path+ " " + s[:i], dp, ind+i, word_dict)
def word_break1(self, s, word_dict):
N = len(s)
dp = [False] * (N+1)
dp[0] =True
for i in range(1, len(s) + 1):
for j in range(i):
if dp[j] and s[j:i] in word_dict:
dp[i] = True
return dp
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
if not s:
return [""]
wordDict = set(wordDict)
dp = self.word_break1(s, wordDict)
print(dp)
self.dfs(s, "", dp, 0, wordDict)
return self.res