146_lru_cache.md

146 - LRU Cache

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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.

Follow up: Could you do get and put in O(1) 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:

• 1 <= capacity <= 3000
• 0 <= key <= 3000
• 0 <= value <= 104
• At most 3 * 104 calls will be made to get and put.

Solution - O(1) time for put & get

// 24/04/2020
class LRUCache {
    int _capacity;
    unordered_map <int, pair<int, list<int>::iterator>> _cache;
    list<int> _lru;
    
public:
    LRUCache(int capacity): _capacity(capacity) {}
    
    int get(int key) {
        auto it = _cache.find(key);
        if (it == _cache.end())
            return -1;
        _lru.splice(_lru.end(), _lru, it->second.second);
        return it->second.first;
    }
    
    void put(int key, int value) {
        auto it = _cache.find(key);
        if (it == _cache.end()){
            if (_cache.size()==_capacity){
                _cache.erase(_lru.front());
                _lru.splice(_lru.end(), _lru, _lru.begin());
                _lru.back()=key;
            }
            else{
                _lru.push_back(key);
            }
            _cache.insert(make_pair(key, make_pair(value, prev(_lru.end()))));
        }else{
            it->second.first = value;
            _lru.splice(_lru.end(), _lru, it->second.second);   
        }
    }
};