Posts published in “Hashtable”

花花酱 LeetCode 460. LFU Cache

By zxi on September 14, 2017

Problem:

Design and implement a data structure for Least Frequently Used (LFU) 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 reaches its capacity, it should invalidate the least frequently used item before inserting a new item. For the purpose of this problem, when there is a tie (i.e., two or more keys that have the same frequency), the least recently used key would be evicted.

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

Example:

LFUCache cache = new LFUCache( 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.get(3); // returns 3.

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:

Hashtable + balanced search tree

Hashtable + double linked list

Solution 1: O(logc) c is the capacity

// Author: Huahua

// Running time: 99 ms

struct CacheNode {

int key;

int value;

int freq;

long tick;

// Defines the order, smaller one will be evicted first

bool operator <(const CacheNode& rhs) const {

if (freq < rhs.freq) return true;

if (freq == rhs.freq) return tick < rhs.tick;

return false;

}

};

class LFUCache {

public:

LFUCache(int capacity):capacity_(capacity), tick_(0) {}

int get(int key) {

auto it = m_.find(key);

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

int value = it->second.value;

touch(it->second);

return value;

}

void put(int key, int value) {

if (capacity_ == 0) return;

auto it = m_.find(key);

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

// Key exists, update value and touch

it->second.value = value;

touch(it->second);

return;

}

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

// Remove the first node in cache

const CacheNode& node = *cache_.cbegin();

m_.erase(node.key);

cache_.erase(node);

}

CacheNode node{key, value, 1, ++tick_};

m_[node.key] = node;

cache_.insert(node);

}

private:

void touch(CacheNode& node) {

cache_.erase(node); // log(capacity)

++node.freq;

node.tick = ++tick_;

cache_.insert(node); // log(capacity)

}

long tick_;

int capacity_;

unordered_map<int, CacheNode> m_;

set<CacheNode> cache_;

};

Solution 2: O(1)

// Author: Huahua

// Running time: 89 ms (beats 96.3%)

struct CacheNode {

int key;

int value;

int freq;

// pointer to the node in the list

list<int>::const_iterator it;

};

class LFUCache {

public:

LFUCache(int capacity): capacity_(capacity), min_freq_(0) {}

int get(int key) {

auto it = n_.find(key);

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

touch(it->second);

return it->second.value;

}

void put(int key, int value) {

if (capacity_ == 0) return;

auto it = n_.find(key);

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

// Key already exists, update the value and touch it

it->second.value = value;

touch(it->second);

return;

}

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

// No capacity, need to remove one entry that

// 1. has the lowest freq

// 2. least recently used if there are multiple ones

// Step 1: remove the element from min_freq_ list

const int key_to_evict = l_[min_freq_].back();

l_[min_freq_].pop_back();

// Step 2: remove the key from cache

n_.erase(key_to_evict);

}

// We know new item has freq of 1, thus set min_freq to 1

const int freq = 1;

min_freq_ = freq;

// Add the key to the freq list

l_[freq].push_front(key);

// Create a new node

n_[key] = {key, value, freq, l_[freq].cbegin()};

}

private:

void touch(CacheNode& node) {

// Step 1: update the frequency

const int prev_freq = node.freq;

const int freq = ++(node.freq);

// Step 2: remove the entry from old freq list

l_[prev_freq].erase(node.it);

if (l_[prev_freq].empty() && prev_freq == min_freq_) {

// Delete the list

l_.erase(prev_freq);

// Increase the min freq

++min_freq_;

}

// Step 4: insert the key into the front of the new freq list

l_[freq].push_front(node.key);

// Step 5: update the pointer

node.it = l_[freq].cbegin();

}

int capacity_;

int min_freq_;

// key -> CacheNode

unordered_map<int, CacheNode> n_;

// freq -> keys with freq

unordered_map<int, list<int>> l_;

};

花花酱 LeetCode 451. Sort Characters By Frequency

By zxi on September 13, 2017

Problem:

Given a string, sort it in decreasing order based on the frequency of characters.

Example 1:

Input:
“tree”

Output:
“eert”

Explanation:
‘e’ appears twice while ‘r’ and ‘t’ both appear once.
So ‘e’ must appear before both ‘r’ and ‘t’. Therefore “eetr” is also a valid answer.
Example 2:

Input:
“cccaaa”

Output:
“cccaaa”

Explanation:
Both ‘c’ and ‘a’ appear three times, so “aaaccc” is also a valid answer.
Note that “cacaca” is incorrect, as the same characters must be together.
Example 3:

Input:
“Aabb”

Output:
“bbAa”

Explanation:
“bbaA” is also a valid answer, but “Aabb” is incorrect.
Note that ‘A’ and ‘a’ are treated as two different characters.

Idea:

Counting sort

Solution 1:

Sort the string based on element frequency

// Author: Huahua

// Running time: 289 ms

class Solution {

public:

string frequencySort(string& s) {

array<int, 128> f{0};

for(char c:s) ++f[c];

sort(s.begin(), s.end(), [&f] (char a, char b) {

return f[a] > f[b] || (f[a] == f[b] && a > b);

});

return s;

}

};

// Author: Huahua

// Running time: 139 ms

class Solution {

public:

string frequencySort(string& s) {

map<char, int> f;

for(char c:s) ++f[c];

sort(s.begin(), s.end(), [&f] (char a, char b) {

return f[a] > f[b] || (f[a] == f[b] && a > b);

});

return s;

}

};

// Author: Huahua

// Running time: 29 ms

class Solution {

public:

string frequencySort(string& s) {

array<int, 128> f{0};

for(char c:s) ++f[c];

sort(s.begin(), s.end(), [&f] (char a, char b) {

return f[a] > f[b] || (f[a] == f[b] && a > b);

});

return s;

}

};

Solution 2:

Counting sort

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

string frequencySort(const string& s) {

vector<int> f(128, 0);

for(char c:s) ++f[c];

vector<pair<int,char>> v;

for (int i = 0; i < 128; ++i) {

if (f[i] > 0) v.emplace_back(f[i], i);

}

sort(v.rbegin(), v.rend());

string ans;

for(const auto& kv : v) {

ans.append(kv.first, kv.second);

}

return ans;

}

};

花花酱 LeetCode 146. LRU Cache O(1)

By zxi on September 10, 2017

Problem:

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

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);

花花酱 LeetCode 676. Implement Magic Dictionary

By zxi on September 10, 2017

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;

}