Problem: LRU Cache
leetcode.com/problems/lru-cache/solution/

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

• LRUCache(int capacity) Initialize the LRU cache with positive size capacity.
• int get(int key) Return the value of the key if the key exists, otherwise return -1.
• void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key.

Follow up:
Could you do get and put in O(1) time complexity?

 

Example 1:

Input
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, null, -1, 3, 4]

Explanation
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // cache is {1=1}
lRUCache.put(2, 2); // cache is {1=1, 2=2}
lRUCache.get(1);    // return 1
lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2);    // returns -1 (not found)
lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(1);    // return -1 (not found)
lRUCache.get(3);    // return 3
lRUCache.get(4);    // return 4

 

Constraints:

• 1 <= capacity <= 3000
• 0 <= key <= 3000
• 0 <= value <= 104
• At most 3 * 104 calls will be made to get and put.

SOLUTION

Introducing Ordered Dictionary collection

>>> from collections import OrderedDict
>>> d = {1:1, 2:2, 3:3}
>>> d = OrderedDict(d)
>>> d
OrderedDict([(1, 1), (2, 2), (3, 3)])
>>> d.move_to_end(2)
>>> d
OrderedDict([(1, 1), (3, 3), (2, 2)])

the Dictionary can also act as a stack or a queue:
stack - use popitem()
queue - use popitem(last=False)

>>> d
OrderedDict([(1, 1), (3, 3), (2, 2)])
>>> d.popitem()
(2, 2)
>>> d
OrderedDict([(1, 1), (3, 3)])
>>> d.popitem(last=False)
(1, 1)
>>> d
OrderedDict([(3, 3)])

And the final solution is:

from collections import OrderedDict

class LRUCache(OrderedDict):
    capacity = 0
    
    def __init__(self, capacity: int):
        self.capacity = capacity
        
    def get(self, key: int) -> int:
        if key in self:
            self.move_to_end(key)
            return self[key]
        else:
            return -1

    def put(self, key: int, value: int) -> None:
        if key in self:
            self.move_to_end(key)
        self[key] = value
        if len(self) > self.capacity:
            self.popitem(last=False)