LeetCode - LRU Cache

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.

The functions get and put must each run in O(1) average 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 <= 104
• 0 <= value <= 105
• At most 2 * 105 calls will be made to get and put.

 

My Solution:

class LRUCache(object):

    def __init__(self, capacity):
        """
        :type capacity: int
        """

        self.capacity = capacity
        self._cache = {}
        self._used = []
        

    def get(self, key):
        """
        :type key: int
        :rtype: int
        """

        try:
            res = self._cache[key]
            self._used.remove(key)
            self._used.append(key)
        except KeyError:
            return -1

        return res
        

    def put(self, key, value):
        """
        :type key: int
        :type value: int
        :rtype: None
        """

        self._cache[key] = value
        if key in self._used:
            self._used.remove(key)
        self._used.append(key)
        if len(self._cache) > self.capacity:
            del self._cache[self._used[0]]
            self._used.pop(0)



# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)

 

 

ChatGPT's Solution:

from collections import OrderedDict


class LRUCache(object):

    def __init__(self, capacity):
        """
        :type capacity: int
        """

        self.cache = OrderedDict()
        self.capacity = capacity
        

    def get(self, key):
        """
        :type key: int
        :rtype: int
        """

        if key not in self.cache:
            return -1
        
        # Move the accessed item to the end to mark it as recently used
        self.cache.move_to_end(key)
        return self.cache[key]
        

    def put(self, key, value):
        """
        :type key: int
        :type value: int
        :rtype: None
        """

        if key in self.cache:
            # Update the key and move it to the end
            self.cache.move_to_end(key)
        self.cache[key] = value
        if len(self.cache) > self.capacity:
            # Pop the least recently used item (first item)
            self.cache.popitem(last=False)
        


# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)