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Commit e76bd94

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Add easy and hard problems
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"""
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# LONGEST WORD IN DICTIONARY
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Given an array of strings words representing an English Dictionary, return the longest word in words that can be built one character at a time by other words in words.
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If there is more than one possible answer, return the longest word with the smallest lexicographical order. If there is no answer, return the empty string.
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Example 1:
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Input: words = ["w","wo","wor","worl","world"]
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Output: "world"
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Explanation: The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".
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Example 2:
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Input: words = ["a","banana","app","appl","ap","apply","apple"]
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Output: "apple"
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Explanation: Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".
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Constraints:
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1 <= words.length <= 1000
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1 <= words[i].length <= 30
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words[i] consists of lowercase English letters.
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"""
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from typing import List
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class Solution:
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def longestWord(self, words: List[str]) -> str:
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words.sort(key = lambda x: (len(x), x))
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trie = Trie()
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longestWord = (None, 0)
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for word in words:
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flag = trie.insert(word)
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if flag and len(word) > longestWord[1]:
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longestWord = (word, len(word))
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return longestWord[0] if longestWord[0] is not None else ""
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class Trie:
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def __init__(self):
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self.root = {}
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self.endSymbol = "*"
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self.root[self.endSymbol] = True
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def insert(self, word):
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current = self.root
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flag = True
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for letter in word:
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if self.endSymbol not in current:
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flag = False
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if letter not in current:
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current[letter] = {}
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current = current[letter]
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current[self.endSymbol] = True
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return flag
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"""
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# PREFIX AND SUFFIX SEARCH
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Design a special dictionary with some words that searchs the words in it by a prefix and a suffix.
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Implement the WordFilter class:
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WordFilter(string[] words) Initializes the object with the words in the dictionary.
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f(string prefix, string suffix) Returns the index of the word in the dictionary, which has the prefix prefix and the suffix suffix. If there is more than one valid index, return the largest of them. If there is no such word in the dictionary, return -1.
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Example 1:
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Input
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["WordFilter", "f"]
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[[["apple"]], ["a", "e"]]
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Output
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[null, 0]
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Explanation
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WordFilter wordFilter = new WordFilter(["apple"]);
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wordFilter.f("a", "e"); // return 0, because the word at index 0 has prefix = "a" and suffix = 'e".
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Constraints:
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1 <= words.length <= 15000
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1 <= words[i].length <= 10
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1 <= prefix.length, suffix.length <= 10
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words[i], prefix and suffix consist of lower-case English letters only.
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At most 15000 calls will be made to the function f.
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"""
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# THIS SOLUTION WILL RESULT IN TIME LIMIT EXCEEDED ERROR.
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from typing import Collection, List
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class WordFilter:
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def __init__(self, words: List[str]):
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self.root = {}
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self.endSymbol = "*"
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self.suffixIndex = "SuffixIndexes"
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self.prefixIndex = "PrefixIndexes"
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for index, word in enumerate(words):
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self.populateTrie(word, index)
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def populateTrie(self, word, index):
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for i in range(len(word)):
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current = self.root
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for j in range(i, len(word)):
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letter = word[j]
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if letter not in current:
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current[letter] = {}
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current = current[letter]
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if self.endSymbol in current and self.suffixIndex in current:
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current[self.suffixIndex].append(index)
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else:
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current[self.endSymbol] = True
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current[self.suffixIndex] = [index]
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for i in range(len(word)):
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current = self.root
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for j in range(i + 1):
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letter = word[j]
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if letter not in current:
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current[letter] = {}
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current = current[letter]
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if self.endSymbol in current and self.prefixIndex in current:
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current[self.prefixIndex].append(index)
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else:
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current[self.endSymbol] = True
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current[self.prefixIndex] = [index]
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def f(self, prefix: str, suffix: str) -> int:
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prefixIndexes = self.searchPrefixIndexes(prefix)
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suffixIndexes = self.searchSuffixIndexes(suffix)
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if prefixIndexes == -1 or suffixIndexes == -1:
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return -1
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prefixIndexes.sort(reverse=True)
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suffixIndexes.sort(reverse=True)
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i, j = 0, 0
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while i < len(prefixIndexes) and j < len(suffixIndexes):
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if prefixIndexes[i] == suffixIndexes[j]:
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return prefixIndexes[i]
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elif prefixIndexes[i] > suffixIndexes[j]:
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i += 1
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else:
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j += 1
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return -1
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def searchPrefixIndexes(self, prefix):
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current = self.root
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for letter in prefix:
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if letter not in current:
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return -1
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current = current[letter]
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if self.endSymbol in current:
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return current[self.prefixIndex]
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def searchSuffixIndexes(self, suffix):
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current = self.root
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for letter in suffix:
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if letter not in current:
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return -1
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current = current[letter]
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if self.endSymbol in current:
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return current[self.suffixIndex]
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# Your WordFilter object will be instantiated and called as such:
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# obj = WordFilter(words)
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# param_1 = obj.f(prefix,suffix)
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########################################
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# THIS IS THE MOST EFFICIENT SOLUTION
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########################################
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Trie = lambda: Collection.defaultdict(Trie)
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WEIGHT = False
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class WordFilter(object):
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def __init__(self, words):
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self.trie = Trie()
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for weight, word in enumerate(words):
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word += '#'
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for i in range(len(word)):
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cur = self.trie
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cur[WEIGHT] = weight
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for j in range(i, 2 * len(word) - 1):
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cur = cur[word[j % len(word)]]
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cur[WEIGHT] = weight
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def f(self, prefix, suffix):
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cur = self.trie
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for letter in suffix + '#' + prefix:
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if letter not in cur:
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return -1
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cur = cur[letter]
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return cur[WEIGHT]

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