Today I have done leetcode's July LeetCoding Challenge with cpp.

July LeetCoding Challenge 18

Description

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.

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 <= 104
• 0 <= value <= 105
• At most 2 * 105 calls will be made to get and put.

Solution

class LRUCache {
  int capacity;
  list<int> LRU;
  unordered_map<int, int> kv;
  unordered_map<int, list<int>::iterator> position;
  
  void update(int key) {
    LRU.erase(position[key]);
    add(key);
  }
  
  void add(int key) {
    LRU.push_front(key);
    position[key] = LRU.begin();
  }
  
  void evict_last() {
    kv.erase(LRU.back());
    position.erase(LRU.back());
    LRU.pop_back(); 
  }
public:
  LRUCache(int capacity) : capacity(capacity) {}
  
  int get(int key) {
    if(kv.count(key)) {
      update(key);
      return kv[key];
    }
    return -1;
  }
  
  void put(int key, int value) {
    if(kv.count(key)) {
      update(key);
    } else {
      if(LRU.size() == capacity) evict_last();
      add(key);
    }
    kv[key] = value;
  }
};

// Accepted
// 22/22 cases passed (538 ms)
// Your runtime beats 27.06 % of cpp submissions
// Your memory usage beats 41.51 % of cpp submissions (175.8 MB)