LFU Cache

// https://leetcode.com/problems/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.

Ex1:
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

*/

#include <list>
#include <unordered_map>
#include <cassert>

using namespace std;

class DLLNode {
public:
    int key;
    int value;
    DLLNode* prev;
    DLLNode* next;
    DLLNode() {}
    DLLNode(int k, int v): key(k), value(v) {}
};

class DLL {
public:
    DLLNode* head;
    DLLNode* tail;
    DLL() {
        head = new DLLNode();
        tail = new DLLNode();
        head->next = tail;
        tail->prev = head;
    }
};

class LRUCache {
public:
    DLL dll;
    int cap;
    unordered_map<int, DLLNode*> m;
    LRUCache(int capacity) {
        cap = capacity;
    }

    int get(int key) {
        if (!m.count(key)) {
            return -1;
        }

        // Move the node to the beginning of the list.
        remove(m[key]);
        insert(m[key]);

        return m[key]->value;
    }

    void put(int key, int value) {
        if (m.count(key)) {
            m[key]->value = value;
            remove(m[key]);
            insert(m[key]);
            return;
        }

        DLLNode* n = new DLLNode(key, value);
        insert(n);
        m[key] = n;

        // remove the least used node at the very end.
        if (m.size() > cap) {
            DLLNode* last = dll.tail->prev;
            // remove the last one.
            remove(last);
            m.erase(last->key);
        }
        return;
    }

    void remove(DLLNode* node) {
        node->prev->next = node->next;
        node->next->prev = node->prev;
        node->prev = nullptr;
        node->next = nullptr;
    }

    void insert(DLLNode* node) {
        node->next = dll.head->next;
        node->prev = dll.head;
        dll.head->next->prev = node;
        dll.head->next = node;
    }
};

int main() {
    LRUCache lRUCache(2);
    lRUCache.put(1, 1); // cache is {1=1}
    lRUCache.put(2, 2); // cache is {1=1, 2=2}
    assert(lRUCache.get(1) == 1);

    lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
    assert(lRUCache.get(2) == -1);

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

    return 0;
}```