Medium
Synonymous Sentences — Python
Full explanation · Time O(p * l * log(p * l)) · Space O(p * l)
# Time: O(p*l * log(p*l)), p is the production of all number of synonyms
# , l is the length of a word
# Space: O(p*l)
import collections
import itertools
class UnionFind(object):
def __init__(self, n):
self.set = range(n)
self.count = n
def find_set(self, x):
if self.set[x] != x:
self.set[x] = self.find_set(self.set[x]) # path compression.
return self.set[x]
def union_set(self, x, y):
x_root, y_root = map(self.find_set, (x, y))
if x_root == y_root:
return False
self.set[max(x_root, y_root)] = min(x_root, y_root)
return True
class Solution(object):
def generateSentences(self, synonyms, text):
"""
:type synonyms: List[List[str]]
:type text: str
:rtype: List[str]
"""
def assign_id(x, lookup, inv_lookup):
if x in lookup:
return
lookup[x] = len(lookup)
inv_lookup[lookup[x]] = x
lookup, inv_lookup = {}, {}
for u, v in synonyms:
assign_id(u, lookup, inv_lookup), assign_id(v, lookup, inv_lookup)
union_find = UnionFind(len(lookup))
for u, v in synonyms:
union_find.union_set(lookup[u], lookup[v])
groups = collections.defaultdict(list)
for i in xrange(len(union_find.set)):
groups[union_find.find_set(i)].append(i)
result = []
for w in text.split(' '):
if w not in lookup:
result.append([w])
continue
result.append(sorted(map(lambda x: inv_lookup[x],
groups[union_find.find_set(lookup[w])])))
return [" ".join(sentense) for sentense in itertools.product(*result)]