Medium
Word Squares II — Python
Full explanation · Time O(n^4) · Space O(1)
# Time: O(n^4)
# Space: O(n)
import collections
# sort, brute force, hash table
class Solution(object):
def wordSquares(self, words):
"""
:type words: List[str]
:rtype: List[List[str]]
"""
words.sort()
lookup = collections.defaultdict(list)
for i, w in enumerate(words):
lookup[w[0]].append(i)
lookup[w[0], w[3]].append(i)
result = []
for i in xrange(len(words)):
for j in lookup[words[i][0]]:
if j == i:
continue
for k in lookup[words[i][3]]:
if k in (i, j):
continue
for l in lookup[words[j][3], words[k][3]]:
if l in (i, j, k):
continue
result.append([words[i], words[j], words[k], words[l]])
return result
# Time: O(n^4)
# Space: O(1)
# sort, brute force
class Solution2(object):
def wordSquares(self, words):
"""
:type words: List[str]
:rtype: List[List[str]]
"""
words.sort()
result = []
for i in xrange(len(words)):
for j in xrange(len(words)):
if j == i or words[j][0] != words[i][0]:
continue
for k in xrange(len(words)):
if k in (i, j) or words[k][0] != words[i][3]:
continue
for l in xrange(len(words)):
if l in (i, j, k) or words[l][0] != words[j][3] or words[l][3] != words[k][3]:
continue
result.append([words[i], words[j], words[k], words[l]])
return result