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index-pairs-of-a-string.py
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index-pairs-of-a-string.py
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# Time: O(n + m + z), n is the total size of patterns
# , m is the total size of query string
# , z is the number of all matched strings
# Space: O(t), t is the total size of ac automata trie
import collections
class AhoNode(object):
def __init__(self):
self.children = collections.defaultdict(AhoNode)
self.indices = []
self.suffix = None
self.output = None
class AhoTrie(object):
def step(self, letter):
while self.__node and letter not in self.__node.children:
self.__node = self.__node.suffix
self.__node = self.__node.children[letter] if self.__node else self.__root
return self.__get_ac_node_outputs(self.__node)
def __init__(self, patterns):
self.__root = self.__create_ac_trie(patterns)
self.__node = self.__create_ac_suffix_and_output_links(self.__root)
def __create_ac_trie(self, patterns): # Time: O(n), Space: O(t)
root = AhoNode()
for i, pattern in enumerate(patterns):
node = root
for c in pattern:
node = node.children[c]
node.indices.append(i)
return root
def __create_ac_suffix_and_output_links(self, root): # Time: O(n), Space: O(t)
queue = collections.deque()
for node in root.children.itervalues():
queue.append(node)
node.suffix = root
while queue:
node = queue.popleft()
for c, child in node.children.iteritems():
queue.append(child)
suffix = node.suffix
while suffix and c not in suffix.children:
suffix = suffix.suffix
child.suffix = suffix.children[c] if suffix else root
child.output = child.suffix if child.suffix.indices else child.suffix.output
return root
def __get_ac_node_outputs(self, node): # Time: O(z)
result = []
for i in node.indices:
result.append(i)
output = node.output
while output:
for i in output.indices:
result.append(i)
output = output.output
return result
class Solution(object):
def indexPairs(self, text, words):
"""
:type text: str
:type words: List[str]
:rtype: List[List[int]]
"""
result = []
reversed_words = [w[::-1] for w in words]
trie = AhoTrie(reversed_words)
for i in reversed(xrange(len(text))):
for j in trie.step(text[i]):
result.append([i, i+len(reversed_words[j])-1])
result.reverse()
return result