LRU Cache
<p>Design a data structure that follows the constraints of a <strong><a href="https://en.wikipedia.org/wiki/Cache_replacement_policies#LRU" target="_blank">Least Recently Used (LRU) cache</a></strong>.</p>

<p>Implement the <code>LRUCache</code> class:</p>

<ul>
	<li><code>LRUCache(int capacity)</code> Initialize the LRU cache with <strong>positive</strong> size <code>capacity</code>.</li>
	<li><code>int get(int key)</code> Return the value of the <code>key</code> if the key exists, otherwise return <code>-1</code>.</li>
	<li><code>void put(int key, int value)</code> Update the value of the <code>key</code> if the <code>key</code> exists. Otherwise, add the <code>key-value</code> pair to the cache. If the number of keys exceeds the <code>capacity</code> from this operation, <strong>evict</strong> the least recently used key.</li>
</ul>

<p>The functions <code>get</code> and <code>put</code> must each run in <code>O(1)</code> average time complexity.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input</strong>
[&quot;LRUCache&quot;, &quot;put&quot;, &quot;put&quot;, &quot;get&quot;, &quot;put&quot;, &quot;get&quot;, &quot;put&quot;, &quot;get&quot;, &quot;get&quot;, &quot;get&quot;]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
<strong>Output</strong>
[null, null, null, 1, null, -1, null, -1, 3, 4]

<strong>Explanation</strong>
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
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= capacity &lt;= 3000</code></li>
	<li><code>0 &lt;= key &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= value &lt;= 10<sup>5</sup></code></li>
	<li>At most <code>2 * 10<sup>5</sup></code> calls will be made to <code>get</code> and <code>put</code>.</li>
</ul>
Design; Hash Table; Linked List; Doubly-Linked List
class Node:
    def __init__(self, key: int = 0, val: int = 0):
        self.key = key
        self.val = val
        self.prev = None
        self.next = None


class LRUCache:
    def __init__(self, capacity: int):
        self.size = 0
        self.capacity = capacity
        self.cache = {}
        self.head = Node()
        self.tail = Node()
        self.head.next = self.tail
        self.tail.prev = self.head

    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1
        node = self.cache[key]
        self.remove_node(node)
        self.add_to_head(node)
        return node.val

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            node = self.cache[key]
            self.remove_node(node)
            node.val = value
            self.add_to_head(node)
        else:
            node = Node(key, value)
            self.cache[key] = node
            self.add_to_head(node)
            self.size += 1
            if self.size > self.capacity:
                node = self.tail.prev
                self.cache.pop(node.key)
                self.remove_node(node)
                self.size -= 1

    def remove_node(self, node):
        node.prev.next = node.next
        node.next.prev = node.prev

    def add_to_head(self, node):
        node.next = self.head.next
        node.prev = self.head
        self.head.next = node
        node.next.prev = node


# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)
