# LRU Cache

Published: Sep 9, 2022

Medium Design Hash Table

## Problem Description

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.

Constraints:

• `1 <= capacity <= 3000`
• `0 <= key <= 10**4`
• `0 <= value <= 10**5`
• At most `2 * 10**5` calls will be made to get and put.

https://leetcode.com/problems/lru-cache/

## Examples

``````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.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(3);    // return 3
lRUCache.get(4);    // return 4
``````

## How to Solve

This is well-known LRU (least recently used) cache problem. The cache size and key order are what we should maintain. What data structure to save those is the key to solve the problem. The data structure should guarantee the order of keys in first come basis. The Python solution uses OrderedDict. Ruby’s hash table keeps the order of addition, so the Ruby solution uses a Hash simply. C++ has a doubly linked list in the standard library. The C++ solution uses it along with an unordered map.

Everytime the key is referenced, that key moved to the end (Python and Ruby) or head (C++) for both `get/put` method. When the cache size grows up to the specified capacity, the least recently used key is deleted.

## Solution

``````class LRUCache {
private:
int capacity;
int n = 0;
unordered_map<int, list<pair<int,int>>::iterator> key2list;
list<pair<int,int>> cache;

public:
LRUCache(int capacity) : capacity(capacity) {}

int get(int key) {
if (key2list.find(key) == key2list.end()) { return -1; }
auto value = key2list[key];
cache.splice(cache.begin(),cache, value);
return key2list[key]->second;
}

void put(int key, int value) {
if (capacity <= 0) { return; }
if (key2list.find(key) != key2list.end()) {
auto itr = key2list[key];
itr->second = value;
cache.splice(cache.begin(), cache, itr);
return;
} else if (n == capacity) {
auto key_to_delete = cache.back().first;
key2list.erase(key_to_delete);
cache.pop_back();
n--;
}
n++;
cache.emplace_front(key,value);
key2list[key] = cache.begin();
}
};
``````
``````
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``````
``````class LRUCache:

def __init__(self, capacity: int):
self.cache = collections.OrderedDict()
self.capacity = capacity

def get(self, key: int) -> int:
if key in self.cache:
self.cache.move_to_end(key)
return self.cache[key]
else:
return -1

def put(self, key: int, value: int) -> None:
if key not in self.cache and len(self.cache) == self.capacity:
self.cache.popitem(last=False)
self.cache[key] = value
self.cache.move_to_end(key)
``````
``````class LRUCache

=begin
:type capacity: Integer
=end
def initialize(capacity)
@capacity = capacity
@cache = {}
end

=begin
:type key: Integer
:rtype: Integer
=end
def get(key)
return -1 if !@cache.has_key?(key)
value = @cache.delete(key)
@cache[key] = value
value
end

=begin
:type key: Integer
:type value: Integer
:rtype: Void
=end
def put(key, value)
if @cache.has_key?(key)
@cache.delete(key)
elsif @cache.size == @capacity
@cache.shift
end
@cache[key] = value
nil
end
end
``````

## Complexities

• Time: `O(1)`
• Space: `O(n)` – n is a capacity