prefix-and-suffix-search.py

# Time:  ctor:   O(w * l^2), w is the number of words, l is the word length on average
#        search: O(p + s)  , p is the length of the prefix, s is the length of the suffix,
# Space: O(t), t is the number of trie nodes

# Given many words, words[i] has weight i.
#
# Design a class WordFilter that supports one function,
# WordFilter.f(String prefix, String suffix).
# It will return the word with given prefix and suffix with maximum weight.
# If no word exists, return -1.
#
# Examples:
# Input:
# WordFilter(["apple"])
# WordFilter.f("a", "e") // returns 0
# WordFilter.f("b", "") // returns -1
# Note:
# words has length in range [1, 15000].
# For each test case, up to words.length queries WordFilter.f may be made.
# words[i] has length in range [1, 10].
# prefix, suffix have lengths in range [0, 10].
# words[i] and prefix, suffix queries consist of lowercase letters only.

import collections


class WordFilter(object):

    def __init__(self, words):
        """
        :type words: List[str]
        """
        _trie = lambda: collections.defaultdict(_trie)
        self.__trie = _trie()

        for weight, word in enumerate(words):
            word += '#'
            for i in xrange(len(word)):
                cur = self.__trie
                cur["_weight"] = weight
                for j in xrange(i, 2*len(word)-1):
                    cur = cur[word[j%len(word)]]
                    cur["_weight"] = weight

    def f(self, prefix, suffix):
        """
        :type prefix: str
        :type suffix: str
        :rtype: int
        """
        cur = self.__trie
        for letter in suffix + '#' + prefix:
            if letter not in cur:
                return -1
            cur = cur[letter]
        return cur["_weight"]


# Time:  ctor:   O(w * l), w is the number of words, l is the word length on average
#        search: O(p + s + max(m, n)), p is the length of the prefix, s is the length of the suffix,
#                                      m is the number of the prefix match, n is the number of the suffix match
# Space: O(w * l)
class Trie(object):

    def __init__(self):
        _trie = lambda: collections.defaultdict(_trie)
        self.__trie = _trie()

    def insert(self, word, i):
        def add_word(cur, i):
            if "_words" not in cur:
                cur["_words"] = []
            cur["_words"].append(i)

        cur = self.__trie
        add_word(cur, i)
        for c in word:
            cur = cur[c]
            add_word(cur, i)

    def find(self, word):
        cur = self.__trie
        for c in word:
            if c not in cur:
                return []
            cur = cur[c]
        return cur["_words"]


class WordFilter2(object):

    def __init__(self, words):
        """
        :type words: List[str]
        """
        self.__prefix_trie = Trie()
        self.__suffix_trie = Trie()
        for i in reversed(xrange(len(words))):
            self.__prefix_trie.insert(words[i], i)
            self.__suffix_trie.insert(words[i][::-1], i)

    def f(self, prefix, suffix):
        """
        :type prefix: str
        :type suffix: str
        :rtype: int
        """
        prefix_match = self.__prefix_trie.find(prefix)
        suffix_match = self.__suffix_trie.find(suffix[::-1])
        i, j = 0, 0
        while i != len(prefix_match) and j != len(suffix_match):
            if prefix_match[i] == suffix_match[j]:
                return prefix_match[i]
            elif prefix_match[i] > suffix_match[j]:
                i += 1
            else:
                j += 1
        return -1


# Your WordFilter object will be instantiated and called as such:
# obj = WordFilter(words)
# param_1 = obj.f(prefix,suffix)