LRUCache.java

/*
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.
*/ 

class LRUCache {

    private final int capacity;
    private final LinkedHashMap<Integer, Integer> map;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        // accessOrder = true for LRU order
        this.map = new LinkedHashMap<>(capacity, 0.75f, true);
    }

    public int get(int key) {
        return map.getOrDefault(key, -1);
    }

    public void put(int key, int value) {
        if (map.containsKey(key)) {
            map.put(key, value); // update value and reorder
        } else {
            if (map.size() >= capacity) {
                // Remove the least recently used entry
                int lruKey = map.keySet().iterator().next(); // Gets the least recently used key (first in access-order LinkedHashMap)
                // .keySet() gives me all key as a set same as in stored in map orderly
                // .iterator() return a object to iterate over the keys
                // .next() returns the first key in the keySet which is least recently used
                map.remove(lruKey);
            }
            map.put(key, value);
        }
    }
}