Word Break
Created: October 10, 2018 by [lek-tin]
Last updated: October 10, 2018
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
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 = "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
Solution (Dynamic programming)
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
if not wordDict:
return False
n = len(s)
flags = [True] + [False] * n
# leetcode
# ["leet", "code"]
# flags[0] = flags[4] = flags[8] = True
for i in range(1, n+1):
for j in range(i):
substring = s[j:i]
print(substring)
if flags[j] and substring in wordDict:
# can break at (i-1)
print(j, i)
flags[i] = True
break
print("-------")
return flags[n]
Solution (BFS)
from collections import deque
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
if not wordDict:
return False
n = len(s)
visited = [False for i in range(n)]
q = deque([])
q.append(0)
while len(q) > 0:
start = q.popleft()
if not visited[start]:
for end in range(start+1, n+1):
substring = s[start:end]
if substring in wordDict:
q.append(end)
# reach the end, return True
if end == n:
return True
# mark start as visited
visited[start] = True
# Not found
return False