<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 <code data-stringify-type="code">get</code> and <code data-stringify-type="code">put</code> must each run in <code>O(1)</code> average time complexity.</p>
<p> </p>
<p><strong>Example 1:</strong></p>
<pre>
<strong>Input</strong>
["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"]
[[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.
// 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> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>0 <= capacity <= 10<sup>4</sup></code></li>
<li><code>0 <= key <= 10<sup>5</sup></code></li>
<li><code>0 <= value <= 10<sup>9</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>
<p> </p>
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