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replace-words.py
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replace-words.py
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# Time: O(n)
# Space: O(t), t is the number of nodes in trie
# In English, we have a concept called root, which can be followed by
# some other words to form another longer word - let's call this word successor.
# For example, the root an, followed by other, which can form another word another.
#
# Now, given a dictionary consisting of many roots and a sentence.
# You need to replace all the successor in the sentence with the root forming it.
# If a successor has many roots can form it, replace it with the root with the shortest length.
#
# You need to output the sentence after the replacement.
#
# Example 1:
# Input: dict = ["cat", "bat", "rat"]
# sentence = "the cattle was rattled by the battery"
# Output: "the cat was rat by the bat"
# Note:
# The input will only have lower-case letters.
# 1 <= dict words number <= 1000
# 1 <= sentence words number <= 1000
# 1 <= root length <= 100
# 1 <= sentence words length <= 1000
class Solution(object):
def replaceWords(self, dictionary, sentence):
"""
:type dictionary: List[str]
:type sentence: str
:rtype: str
"""
_trie = lambda: collections.defaultdict(_trie)
trie = _trie()
for word in dictionary:
reduce(dict.__getitem__, word, trie).setdefault("_end")
def replace(word):
curr = trie
for i, c in enumerate(word):
if c not in curr:
break
curr = curr[c]
if "_end" in curr:
return word[:i+1]
return word
return " ".join(map(replace, sentence.split()))