12. LRU Cache (V.IMP)

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.

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

Solution: (Using Hashmap and linked list stl)

class LRUCache
{
public:
    list<pair<int, int>> ll;
    unordered_map<int, list<pair<int, int>>::iterator> m;
    int cap, count;

    LRUCache(int capacity)
    {
        count = 0;
        cap = capacity;
    }

    int get(int key)
    {
        if (m.find(key) != m.end())
        {
            list<pair<int, int>>::iterator itr;
            itr = m[key];

            //Refresh Cache
            ll.push_front({key, (*itr).second});
            ll.erase(itr);
            m[key] = ll.begin();

            return ll.front().second;
        }

        return -1;
    }

    void put(int key, int value)
    {
        if (m.find(key) != m.end())
        {
            list<pair<int, int>>::iterator itr;
            itr = m[key];
            
            //Refresh Cache
            ll.push_front({key, value});
            m[key] = ll.begin();
            ll.erase(itr);
        }
        else
        {
            //Insert new value
            ll.push_front({key, value});
            m[key] = ll.begin();
            count++;
            
            //Check size and erase old
            if (count > cap)
            {
                m.erase(ll.back().first);
                ll.pop_back();
                count--;
            }
        }

        return;
    }
};

Time Complexity: Get : O(1), Put : O(1)