LFU Cache
<p>Design and implement a data structure for a <a href="https://en.wikipedia.org/wiki/Least_frequently_used" target="_blank">Least Frequently Used (LFU)</a> cache.</p>

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

<ul>
	<li><code>LFUCache(int capacity)</code> Initializes the object with the <code>capacity</code> of the data structure.</li>
	<li><code>int get(int key)</code> Gets the value of the <code>key</code> if the <code>key</code> exists in the cache. Otherwise, returns <code>-1</code>.</li>
	<li><code>void put(int key, int value)</code> Update the value of the <code>key</code> if present, or inserts the <code>key</code> if not already present. When the cache reaches its <code>capacity</code>, it should invalidate and remove the <strong>least frequently used</strong> key before inserting a new item. For this problem, when there is a <strong>tie</strong> (i.e., two or more keys with the same frequency), the <strong>least recently used</strong> <code>key</code> would be invalidated.</li>
</ul>

<p>To determine the least frequently used key, a <strong>use counter</strong> is maintained for each key in the cache. The key with the smallest <strong>use counter</strong> is the least frequently used key.</p>

<p>When a key is first inserted into the cache, its <strong>use counter</strong> is set to <code>1</code> (due to the <code>put</code> operation). The <strong>use counter</strong> for a key in the cache is incremented either a <code>get</code> or <code>put</code> operation is called on it.</p>

<p>The functions&nbsp;<code data-stringify-type="code">get</code>&nbsp;and&nbsp;<code data-stringify-type="code">put</code>&nbsp;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;LFUCache&quot;, &quot;put&quot;, &quot;put&quot;, &quot;get&quot;, &quot;put&quot;, &quot;get&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], [3], [4, 4], [1], [3], [4]]
<strong>Output</strong>
[null, null, null, 1, null, -1, 3, null, -1, 3, 4]

<strong>Explanation</strong>
// cnt(x) = the use counter for key x
// cache=[] will show the last used order for tiebreakers (leftmost element is  most recent)
LFUCache lfu = new LFUCache(2);
lfu.put(1, 1);   // cache=[1,_], cnt(1)=1
lfu.put(2, 2);   // cache=[2,1], cnt(2)=1, cnt(1)=1
lfu.get(1);      // return 1
                 // cache=[1,2], cnt(2)=1, cnt(1)=2
lfu.put(3, 3);   // 2 is the LFU key because cnt(2)=1 is the smallest, invalidate 2.
&nbsp;                // cache=[3,1], cnt(3)=1, cnt(1)=2
lfu.get(2);      // return -1 (not found)
lfu.get(3);      // return 3
                 // cache=[3,1], cnt(3)=2, cnt(1)=2
lfu.put(4, 4);   // Both 1 and 3 have the same cnt, but 1 is LRU, invalidate 1.
                 // cache=[4,3], cnt(4)=1, cnt(3)=2
lfu.get(1);      // return -1 (not found)
lfu.get(3);      // return 3
                 // cache=[3,4], cnt(4)=1, cnt(3)=3
lfu.get(4);      // return 4
                 // cache=[4,3], cnt(4)=2, cnt(3)=3
</pre>

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

<ul>
	<li><code>1 &lt;= capacity&nbsp;&lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= key &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= value &lt;= 10<sup>9</sup></code></li>
	<li>At most <code>2 * 10<sup>5</sup></code>&nbsp;calls will be made to <code>get</code> and <code>put</code>.</li>
</ul>

<p>&nbsp;</p>
<span style="display: none;">&nbsp;</span>
Design; Hash Table; Linked List; Doubly-Linked List
type LFUCache struct {
	cache    map[int]*node
	freqMap  map[int]*list
	minFreq  int
	capacity int
}

func Constructor(capacity int) LFUCache {
	return LFUCache{
		cache:    make(map[int]*node),
		freqMap:  make(map[int]*list),
		capacity: capacity,
	}
}

func (this *LFUCache) Get(key int) int {
	if this.capacity == 0 {
		return -1
	}

	n, ok := this.cache[key]
	if !ok {
		return -1
	}

	this.incrFreq(n)
	return n.val
}

func (this *LFUCache) Put(key int, value int) {
	if this.capacity == 0 {
		return
	}

	n, ok := this.cache[key]
	if ok {
		n.val = value
		this.incrFreq(n)
		return
	}

	if len(this.cache) == this.capacity {
		l := this.freqMap[this.minFreq]
		delete(this.cache, l.removeBack().key)
	}
	n = &node{key: key, val: value, freq: 1}
	this.addNode(n)
	this.cache[key] = n
	this.minFreq = 1
}

func (this *LFUCache) incrFreq(n *node) {
	l := this.freqMap[n.freq]
	l.remove(n)
	if l.empty() {
		delete(this.freqMap, n.freq)
		if n.freq == this.minFreq {
			this.minFreq++
		}
	}
	n.freq++
	this.addNode(n)
}

func (this *LFUCache) addNode(n *node) {
	l, ok := this.freqMap[n.freq]
	if !ok {
		l = newList()
		this.freqMap[n.freq] = l
	}
	l.pushFront(n)
}

type node struct {
	key  int
	val  int
	freq int
	prev *node
	next *node
}

type list struct {
	head *node
	tail *node
}

func newList() *list {
	head := new(node)
	tail := new(node)
	head.next = tail
	tail.prev = head
	return &list{
		head: head,
		tail: tail,
	}
}

func (l *list) pushFront(n *node) {
	n.prev = l.head
	n.next = l.head.next
	l.head.next.prev = n
	l.head.next = n
}

func (l *list) remove(n *node) {
	n.prev.next = n.next
	n.next.prev = n.prev
	n.next = nil
	n.prev = nil
}

func (l *list) removeBack() *node {
	n := l.tail.prev
	l.remove(n)
	return n
}

func (l *list) empty() bool {
	return l.head.next == l.tail
}