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.
Note that the number of times an item is used is the number of calls to the get and put functions for that item since it was inserted. This number is set to zero when the item is removed.
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
class Node:
def __init__(self, key, value):
self.freq = 1
self.key = key
self.value = value
self.next = None
self.prev = None
class DoubleLinkedList:
def __init__(self):
self._head = Node(0, 0)
self._tail = Node(0, 0)
self._head.next = self._tail
self._tail.prev = self._head
self._size = 0
def append(self, node):
self._connect(node, self._head.next)
self._connect(self._head, node)
self._size += 1
def pop(self, node):
self._connect(node.prev, node.next)
node.next = None
node.prev = None
self._size -= 1
return node
def pop_last(self):
return self.pop(self._tail.prev)
def get_size(self):
return self._size
def _connect(self, p, n):
p.next, n.prev = n, p
class LFUCache:
def __init__(self, capacity):
self.nodes = {}
self.freq = {}
self.min_freq = 1
self.capacity = capacity
def get(self, key):
if key in self.nodes:
node = self.nodes[key]
l = self.freq[node.freq]
l.pop(node)
if l.get_size() == 0 and self.min_freq == node.freq:
self.min_freq += 1
node.freq += 1
if self.freq.get(node.freq):
self.freq[node.freq].append(node)
else:
l = DoubleLinkedList()
l.append(node)
self.freq[node.freq] = l
return node.value
else:
return -1
def put(self, key, value):
if self.capacity == 0:
return
if key in self.nodes:
self.nodes[key].value = value
self.get(key)
else:
node = Node(key, value)
if len(self.nodes) == self.capacity:
l = self.freq[self.min_freq]
n = l.pop_last()
self.nodes.pop(n.key)
self.nodes[key] = node
if 1 in self.freq:
self.freq[1].append(node)
else:
l = DoubleLinkedList()
l.append(node)
self.freq[1] = l
self.min_freq = 1
c = LFUCache(3)
c.put(1, 1)
c.put(2, 2)
print(c.get(1))
print(c.get(2))
print(c.get(1))
c.put(3, 3)
print(c.get(2))
print(c.get(3))
c.put(4, 4)
print(c.get(4))