LRU Cache

LRU Cache - LeetCodeLeetCode

Problem

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

Pseudocode

- implement LRU from scratch ala 
    - https://github.com/nightisyang/kata-machine/blob/main/src/day2/LRU.ts

- or use a combination of set and map,
    - leveraging on the properties of how elements are added in set
    - the first element is the last priority
    - delete and add key (as element) to push to back of Set, queue

Solution

const LRUCache = function (capacity) {
  this.capacity = capacity;
  this.cache = new Set();
  this.obj = new Map();
};

LRUCache.prototype.get = function (key) {
  if (!this.cache.has(key)) {
    return -1;
  } else {
    this.cache.delete(key);
    this.cache.add(key);
  }
  return this.obj.get(key);
};

LRUCache.prototype.put = function (key, value) {
  if (!this.cache.has(key)) {
    this.cache.add(key);
  } else {
    this.cache.delete(key);
    this.cache.add(key);
  }
  if (this.cache.size > this.capacity) {
    let [first] = this.cache;
    this.obj.delete(first);
    this.cache.delete(first);
  }
  this.obj.set(key, value);
};

Time and Space Complexity

Time

• What did the code do

• Total -

Space

• What did the code do

• Total -