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## LRU, LFU cache with Doubly Linked List

Topic: Data Structure

## LRU Cache | 0 | subheading |

```
class Node:
def __init__(self, k, v):
self.key = k
self.val = v
self.prev = None
self.next = None
class LRUCache:
def __init__(self, capacity):
self.capacity = capacity
self.dic = dict()
self.head = Node(0, 0)
self.tail = Node(0, 0)
self.head.next = self.tail
self.tail.prev = self.head
def get(self, key):
if key in self.dic:
n = self.dic[key]
self._remove(n)
self._add(n)
return n.val
return -1
def put(self, key, value):
if key in self.dic:
self._remove(self.dic[key])
n = Node(key, value)
self._add(n)
self.dic[key] = n
if len(self.dic) > self.capacity:
n = self.head.next
self._remove(n)
del self.dic[n.key]
def _remove(self, node):
p = node.prev
n = node.next
p.next = n
n.prev = p
def _add(self, node):
p = self.tail.prev
p.next = node
self.tail.prev = node
node.prev = p
node.next = self.tail
``` | 0 | code |

## LFU Cache | 0 | subheading |

```
class Node:
def __init__(self, key, val):
self.key = key
self.val = val
self.freq = 1
self.prev = self.next = None
class DLinkedList:
''' An implementation of doubly linked list.
Two APIs provided:
append(node): append the node to the head of the linked list.
pop(node=None): remove the referenced node.
If None is given, remove the one from tail, which is the least recently used.
Both operation, apparently, are in O(1) complexity.
'''
def __init__(self):
self._sentinel = Node(None, None) # dummy node
self._sentinel.next = self._sentinel.prev = self._sentinel
self._size = 0
def __len__(self):
return self._size
def append(self, node):
node.next = self._sentinel.next
node.prev = self._sentinel
node.next.prev = node
self._sentinel.next = node
self._size += 1
def pop(self, node=None):
if self._size == 0:
return
if not node:
node = self._sentinel.prev
node.prev.next = node.next
node.next.prev = node.prev
self._size -= 1
return node
class LFUCache:
def __init__(self, capacity):
'''
:type capacity: int
Three things to maintain:
1. a dict, named as `self._node`, for the reference of all nodes given key.
That is, O(1) time to retrieve node given a key.
2. Each frequency has a doubly linked list, store in `self._freq`, where key
is the frequency, and value is an object of `DLinkedList`
3. The min frequency through all nodes. We can maintain this in O(1) time, taking
advantage of the fact that the frequency can only increment by 1. Use the following
two rules:
Rule 1: Whenever we see the size of the DLinkedList of current min frequency is 0,
the min frequency must increment by 1.
Rule 2: Whenever put in a new (key, value), the min frequency must 1 (the new node)
'''
self._size = 0
self._capacity = capacity
self._node = dict() # key: Node
self._freq = collections.defaultdict(DLinkedList)
self._minfreq = 0
def _update(self, node):
'''
This is a helper function that used in the following two cases:
1. when `get(key)` is called; and
2. when `put(key, value)` is called and the key exists.
The common point of these two cases is that:
1. no new node comes in, and
2. the node is visited one more times -> node.freq changed ->
thus the place of this node will change
The logic of this function is:
1. pop the node from the old DLinkedList (with freq `f`)
2. append the node to new DLinkedList (with freq `f+1`)
3. if old DlinkedList has size 0 and self._minfreq is `f`,
update self._minfreq to `f+1`
All of the above opeartions took O(1) time.
'''
freq = node.freq
self._freq[freq].pop(node)
if self._minfreq == freq and not self._freq[freq]:
self._minfreq += 1
node.freq += 1
freq = node.freq
self._freq[freq].append(node)
def get(self, key):
'''
Through checking self._node[key], we can get the node in O(1) time.
Just performs self._update, then we can return the value of node.
:type key: int
:rtype: int
'''
if key not in self._node:
return -1
node = self._node[key]
self._update(node)
return node.val
def put(self, key, value):
'''
If `key` already exists in self._node, we do the same operations as `get`, except
updating the node.val to new value.
Otherwise, the following logic will be performed
1. if the cache reaches its capacity, pop the least frequently used item. (*)
2. add new node to self._node
3. add new node to the DLinkedList with frequency 1
4. reset self._minfreq to 1
(*) How to pop the least frequently used item? Two facts:
1. we maintain the self._minfreq, the minimum possible frequency in cache.
2. All cache with the same frequency are stored as a DLinkedList, with
recently used order (Always append at head)
Consequence? ==> The tail of the DLinkedList with self._minfreq is the least
recently used one, pop it...
:type key: int
:type value: int
:rtype: void
'''
if self._capacity == 0:
return
if key in self._node:
node = self._node[key]
self._update(node)
node.val = value
else:
if self._size == self._capacity:
node = self._freq[self._minfreq].pop()
del self._node[node.key]
self._size -= 1
node = Node(key, value)
self._node[key] = node
self._freq[1].append(node)
self._minfreq = 1
self._size += 1
``` | 0 | code |