LeetCode-in-All

146. LRU Cache

Medium

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

Implement the LRUCache class:

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:

To solve the “LRU Cache” problem, we need to implement a data structure that efficiently supports the get and put operations while maintaining the least recently used (LRU) policy. We can achieve this using a combination of a dictionary and a doubly linked list.

Steps to Implement the LRU Cache

  1. Initialization:
    • Create a class LRUCache with an initializer that takes the capacity of the cache.
    • Use a dictionary to store the key-value pairs for O(1) access.
    • Use a doubly linked list to maintain the order of access for the LRU policy.
  2. Node Class:
    • Define a nested Node class for the doubly linked list nodes, which includes properties for the key, value, previous node, and next node.
  3. Doubly Linked List:
    • Maintain a dummy head and tail to facilitate easy addition and removal of nodes.
    • Implement methods to add a node right after the head and remove a node from the linked list.
  4. Get Operation:
    • If the key exists in the dictionary, move the corresponding node to the head of the linked list and return its value.
    • If the key does not exist, return -1.
  5. Put Operation:
    • If the key exists, update the value and move the corresponding node to the head of the linked list.
    • If the key does not exist, add a new node to the head of the linked list and update the dictionary.
    • If the cache exceeds its capacity, remove the node just before the tail (the least recently used node) and delete it from the dictionary.

Swift Implementation

Here’s the Swift implementation of the LRUCache class:

class LRUCache {

    private class Node {
        var key: Int
        var value: Int
        var prev: Node?
        var next: Node?
        
        init(_ key: Int, _ value: Int) {
            self.key = key
            self.value = value
        }
    }
    
    private var capacity: Int
    private var dict = [Int: Node]()
    private var head = Node(0, 0)
    private var tail = Node(0, 0)
    
    init(_ capacity: Int) {
        self.capacity = capacity
        head.next = tail
        tail.prev = head
    }
    
    func get(_ key: Int) -> Int {
        if let node = dict[key] {
            moveToHead(node)
            return node.value
        } else {
            return -1
        }
    }
    
    func put(_ key: Int, _ value: Int) {
        if let node = dict[key] {
            node.value = value
            moveToHead(node)
        } else {
            let newNode = Node(key, value)
            dict[key] = newNode
            addNode(newNode)
            if dict.count > capacity {
                if let tail = removeTail() {
                    dict.removeValue(forKey: tail.key)
                }
            }
        }
    }
    
    private func addNode(_ node: Node) {
        node.prev = head
        node.next = head.next
        head.next?.prev = node
        head.next = node
    }
    
    private func removeNode(_ node: Node) {
        let prev = node.prev
        let next = node.next
        prev?.next = next
        next?.prev = prev
    }
    
    private func moveToHead(_ node: Node) {
        removeNode(node)
        addNode(node)
    }
    
    private func removeTail() -> Node? {
        let node = tail.prev
        if let node = node {
            removeNode(node)
        }
        return node
    }
}

Explanation of the Swift Code

  1. Node Class:
    • The Node class represents each node in the doubly linked list, with key, value, prev, and next properties.
  2. LRUCache Initialization:
    • The initializer sets the capacity and initializes the dummy head and tail nodes to facilitate easy node additions and removals.
  3. Add and Remove Node:
    • addNode(_:) inserts a node right after the head.
    • removeNode(_:) removes a node from its current position.
    • moveToHead(_:) removes a node and then adds it right after the head.
    • removeTail() removes the node just before the tail, which is the least recently used node.
  4. Get Operation:
    • The get(_:) method checks if the key exists in the dictionary. If it does, it moves the node to the head and returns its value. Otherwise, it returns -1.
  5. Put Operation:
    • The put(_:) method updates the node’s value if the key exists and moves it to the head.
    • If the key does not exist, it creates a new node, adds it to the head, and updates the dictionary.
    • If the cache exceeds its capacity, it removes the least recently used node (node just before the tail) and deletes it from the dictionary.

This implementation ensures that both get and put operations run in O(1) time complexity, adhering to the problem constraints.