Hashtable Archives - Page 34 of 34

Problem:

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) – Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) – Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);

cache.put(2, 2);

cache.get(1); // returns 1

cache.put(3, 3); // evicts key 2

cache.get(2); // returns -1 (not found)

cache.put(4, 4); // evicts key 1

cache.get(1); // returns -1 (not found)

cache.get(3); // returns 3

cache.get(4); // returns 4

Idea:

Use Hashtable for fast mapping and double linked list for fast manipulation.

Time Complexity:

Put O(1)

Get O(1)

Solution:

// Author: Huahua

class LRUCache {

public:

LRUCache(int capacity) {

capacity_ = capacity;

}

int get(int key) {

const auto it = m_.find(key);

// If key does not exist

if (it == m_.cend()) return -1;

// Move this key to the front of the cache

cache_.splice(cache_.begin(), cache_, it->second);

return it->second->second;

}

void put(int key, int value) {

const auto it = m_.find(key);

// Key already exists

if (it != m_.cend()) {

// Update the value

it->second->second = value;

// Move this entry to the front of the cache

cache_.splice(cache_.begin(), cache_, it->second);

return;

}

// Reached the capacity, remove the oldest entry

if (cache_.size() == capacity_) {

const auto& node = cache_.back();

m_.erase(node.first);

cache_.pop_back();

}

// Insert the entry to the front of the cache and update mapping.

cache_.emplace_front(key, value);

m_[key] = cache_.begin();

}

private:

int capacity_;

list<pair<int,int>> cache_;

unordered_map<int, list<pair<int,int>>::iterator> m_;

};

/**

* Your LRUCache object will be instantiated and called as such:

* LRUCache obj = new LRUCache(capacity);

* int param_1 = obj.get(key);

* obj.put(key,value);

Problem:

Implement a magic directory with buildDict, and search methods.

For the method buildDict, you’ll be given a list of non-repetitive words to build a dictionary.

For the method search, you’ll be given a word, and judge whether if you modify exactly one character into another character in this word, the modified word is in the dictionary you just built.

Example 1:

Input: buildDict(["hello", "leetcode"]), Output: Null

Input: search("hello"), Output: False

Input: search("hhllo"), Output: True

Input: search("hell"), Output: False

Input: search("leetcoded"), Output: False

Note:

You may assume that all the inputs are consist of lowercase letters a-z.
For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
Please remember to RESET your class variables declared in class MagicDictionary, as static/class variables are persisted across multiple test cases. Please see here for more details.

Idea:

Fuzzy match

Time Complexity:

buildDict: O(n*m)

n: numbers of words

m: length of word

search: O(m)

m: length of word

Space Complexity:

O(n*m)

Solution:

// Author: Huahua

class MagicDictionary {

public:

/** Initialize your data structure here. */

MagicDictionary() {

dict_.clear();

}

/** Build a dictionary through a list of words */

void buildDict(vector<string> dict) {

for(string& word: dict) {

for(int i = 0; i < word.length(); ++i) {

char c = word[i];

word[i] = '*';

dict_[word].insert(c);

word[i] = c;

}

/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */

bool search(string word) {

for(int i = 0; i < word.length(); ++i) {

char c = word[i];

word[i] = '*';

if (dict_.count(word)) {

const auto& char_set = dict_[word];

if (!char_set.count(c) || char_set.size() > 1)

return true;

}

word[i] = c;

}

return false;

}

private:

unordered_map<string, unordered_set<char>> dict_;

};

/**

* Your MagicDictionary object will be instantiated and called as such:

* MagicDictionary obj = new MagicDictionary();

* obj.buildDict(dict);

* bool param_2 = obj.search(word);

Java / Trie

class MagicDictionary {

private Trie root;

public MagicDictionary() {

root = new Trie();

}

public void buildDict(String[] dict) {

for (String word : dict) {

Trie curr = root;

for (char c : word.toCharArray()) {

int index = c - 'a';

if (curr.children[index] == null)

curr.children[index] = new Trie();

curr = curr.children[index];

}

curr.ends = true;

}

public boolean search(String word) {

char[] s = word.toCharArray();

for(int i = 0; i < s.length; i++) {

for (char c = 'a'; c <= 'z'; ++c) {

if (s[i] == c) continue;

char tmp = s[i];

s[i] = c;

if (contains(s)) return true;

s[i] = tmp;

}

return false;

}

private boolean contains(char[] word) {

Trie curr = root;

for (int i = 0; i < word.length; ++i) {

curr = curr.children[word[i] - 'a'];

if (curr == null) return false;

}

return curr.ends;

}

class Trie {

private boolean ends;

private Trie[] children;

public Trie() {

children = new Trie[26];

ends = false;

}

Java / Trie v2

class MagicDictionary {

private Trie root;

public MagicDictionary() {

root = new Trie();

}

public void buildDict(String[] dict) {

for (String word : dict) {

Trie curr = root;

for (char c : word.toCharArray()) {

int index = c - 'a';

if (curr.children[index] == null)

curr.children[index] = new Trie();

curr = curr.children[index];

}

curr.ends = true;

}

public boolean search(String word) {

char[] s = word.toCharArray();

Trie prefix = root;

for(int i = 0; i < s.length; i++) {

for (char c = 'a'; c <= 'z'; ++c) {

if (s[i] == c) continue;

char tmp = s[i];

s[i] = c;

if (contains(s, prefix, i)) return true;

s[i] = tmp;

}

prefix = prefix.children[s[i] - 'a'];

if (prefix == null) return false;

}

return false;

}

private boolean contains(char[] word, Trie prefix, int s) {

Trie curr = prefix;

for (int i = s; i < word.length; ++i) {

curr = curr.children[word[i] - 'a'];

if (curr == null) return false;

}

return curr.ends;

}

class Trie {

private boolean ends;

private Trie[] children;

public Trie() {

children = new Trie[26];

ends = false;

}

Posts published in “Hashtable”

花花酱 LeetCode 146. LRU Cache O(1)

花花酱 LeetCode 676. Implement Magic Dictionary