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Commit c2242cb

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Add medium problems
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"""
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# IMPLEMENT MAGIC DICTIONARY
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Design a data structure that is initialized with a list of different words. Provided a string, you should determine if you can change exactly one character in this string to match any word in the data structure.
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Implement the MagicDictionary class:
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MagicDictionary() Initializes the object.
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void buildDict(String[] dictionary) Sets the data structure with an array of distinct strings dictionary.
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bool search(String searchWord) Returns true if you can change exactly one character in searchWord to match any string in the data structure, otherwise returns false.
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Example 1:
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Input
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["MagicDictionary", "buildDict", "search", "search", "search", "search"]
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[[], [["hello", "leetcode"]], ["hello"], ["hhllo"], ["hell"], ["leetcoded"]]
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Output
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[null, null, false, true, false, false]
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Explanation
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MagicDictionary magicDictionary = new MagicDictionary();
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magicDictionary.buildDict(["hello", "leetcode"]);
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magicDictionary.search("hello"); // return False
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magicDictionary.search("hhllo"); // We can change the second 'h' to 'e' to match "hello" so we return True
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magicDictionary.search("hell"); // return False
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magicDictionary.search("leetcoded"); // return False
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Constraints:
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1 <= dictionary.length <= 100
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1 <= dictionary[i].length <= 100
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dictionary[i] consists of only lower-case English letters.
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All the strings in dictionary are distinct.
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1 <= searchWord.length <= 100
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searchWord consists of only lower-case English letters.
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buildDict will be called only once before search.
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At most 100 calls will be made to search.
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"""
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class MagicDictionary:
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def __init__(self):
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"""
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Initialize your data structure here.
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"""
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self.root = {}
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self.endSymbol = "*"
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def buildDict(self, dictionary: List[str]) -> None:
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for word in dictionary:
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current = self.root
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for letter in word:
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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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def search(self, searchWord: str) -> bool:
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return self.searchHelper(searchWord, 0, self.root, 0)
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def searchHelper(self, word, index, trieNode, count):
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if index == len(word):
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return count == 1 and self.endSymbol in trieNode
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letter = word[index]
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if letter not in trieNode and count == 1:
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return False
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for node in trieNode:
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if node == self.endSymbol:
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continue
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nextCount = count if letter == node else count + 1
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if self.searchHelper(word, index + 1, trieNode[node], nextCount):
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return True
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return False
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# Your MagicDictionary object will be instantiated and called as such:
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# obj = MagicDictionary()
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# obj.buildDict(dictionary)
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# param_2 = obj.search(searchWord)

‎Leetcode/medium/replace-words.py‎

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# REPLACE WORDS
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"""
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In English, we have a concept called root, which can be followed by some other word to form another longer word - let's call this word successor. For example, when the root "an" is followed by the successor word "other", we can form a new word "another".
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Given a dictionary consisting of many roots and a sentence consisting of words separated by spaces, replace all the successors in the sentence with the root forming it. If a successor can be replaced by more than one root, replace it with the root that has the shortest length.
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Return the sentence after the replacement.
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Example 1:
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Input: dictionary = ["cat","bat","rat"], sentence = "the cattle was rattled by the battery"
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Output: "the cat was rat by the bat"
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Example 2:
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Input: dictionary = ["a","b","c"], sentence = "aadsfasf absbs bbab cadsfafs"
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Output: "a a b c"
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Example 3:
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Input: dictionary = ["a", "aa", "aaa", "aaaa"], sentence = "a aa a aaaa aaa aaa aaa aaaaaa bbb baba ababa"
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Output: "a a a a a a a a bbb baba a"
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Example 4:
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Input: dictionary = ["catt","cat","bat","rat"], sentence = "the cattle was rattled by the battery"
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Output: "the cat was rat by the bat"
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Example 5:
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Input: dictionary = ["ac","ab"], sentence = "it is abnormal that this solution is accepted"
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Output: "it is ab that this solution is ac"
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Constraints:
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1 <= dictionary.length <= 1000
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1 <= dictionary[i].length <= 100
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dictionary[i] consists of only lower-case letters.
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1 <= sentence.length <= 10^6
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sentence consists of only lower-case letters and spaces.
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The number of words in sentence is in the range [1, 1000]
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The length of each word in sentence is in the range [1, 1000]
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Each two consecutive words in sentence will be separated by exactly one space.
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sentence does not have leading or trailing spaces.
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"""
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class Solution:
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def replaceWords(self, dictionary, sentence):
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trie = Trie(dictionary)
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sentence += " "
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sentenceList = list(sentence)
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finalList = []
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trieNode = trie.root
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index = 0
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while index < len(sentenceList):
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character = sentenceList[index]
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if trie.endSymbol in trieNode:
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finalList.append(" ")
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while sentenceList[index] != " ":
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index += 1
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trieNode = trie.root
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elif character not in trieNode:
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print(character, "not")
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while index < len(sentenceList):
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finalList.append(sentenceList[index])
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if sentenceList[index] == " ":
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break
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index += 1
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trieNode = trie.root
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else:
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trieNode = trieNode[character]
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finalList.append(character)
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index += 1
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return "".join(finalList[:-1])
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class Trie:
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def __init__(self, words):
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self.root = {}
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self.endSymbol = "*"
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self.populateTrie(words)
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def populateTrie(self, words):
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for word in words:
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current = self.root
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for letter in word:
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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] = word

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