Package g0401_0500.s0460_lfu_cache

Class LFUCache

  • public class LFUCache
extends Object
    460 - LFU Cache\. Hard Design and implement a data structure for a [Least Frequently Used (LFU)](https://en.wikipedia.org/wiki/Least_frequently_used) 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=[3,4], 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`.

    • Constructor Detail

      • LFUCache

        public LFUCache​(int capacity)

    • Method Detail

      • get

        public int get​(int key)

      • put

        public void put​(int key,
                int value)