Design and implement a data structure for a Least Frequently Used (LFU) cache.

Implement the `LFUCache` class:

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

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

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

The functions `get` and `put` must each run in `O(1)` average time complexity.

Example 1:

    Input
    ["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]]
    Output
    [null, null, null, 1, null, -1, 3, null, -1, 3, 4]

    Explanation
    // 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

Constraints:

- `0 <= capacity <= 104`
- `0 <= key <= 105`
- `0 <= value <= 109`
- At most `2 * 105` calls will be made to `get` and `put`.