LFU Cache

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

Operations

    
LFUCache(int capacity): Initialize the LFU cache with a positive capacity.
    
get(int key): Return the value of the key if it exists, otherwise return -1.
    
put(int key, int value): 
Update the value of the key if present, or insert the key-value pair if not. 
When the cache reaches its capacity, it should invalidate and remove the least frequently used key before adding a new item. 
If there is a tie for the least frequently used key, the least recently used key among them should be removed.
  

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: 

      
Step 1: Initialize cache with capacity = 2.
      
Step 2: cache.put(1, 1): cache=[1,_], cnt(1)=1
      
Step 3: cache.put(2, 2): cache=[1,2], cnt(1)=1, cnt(2)=1
      
Step 4: cache.get(1): return 1, cache=[1,2], cnt(1)=2, cnt(2)=1
      
Step 5: cache.put(3, 3): cache=[1,3], cnt(1)=2, cnt(3)=1, 2 is evicted
      
Step 6: cache.get(2): return -1 (not found)
      
Step 7: cache.get(3): return 3, cache=[1,3], cnt(1)=2, cnt(3)=2
      
Step 8: cache.put(4, 4): cache=[4,3], cnt(4)=1, cnt(3)=2, 1 is evicted
      
Step 9: cache.get(1): return -1 (not found)
      
Step 10: cache.get(3): return 3, cache=[4,3], cnt(4)=1, cnt(3)=3
      
Step 11: cache.get(4): return 4, cache=[4,3], cnt(4)=2, cnt(3)=3
    
  

Example 2

    
Input: ["LFUCache", "put", "put", "put", "get", "get"]
[[2], [1, 10], [2, 20], [3, 30], [1], [2]]
    
Output: [null, null, null, null, -1, 20]
    
Explanation: 

      
Step 1: Initialize cache with capacity = 2.
      
Step 2: cache.put(1, 10): cache=[1,_], cnt(1)=1
      
Step 3: cache.put(2, 20): cache=[1,2], cnt(1)=1, cnt(2)=1
      
Step 4: cache.put(3, 30): cache=[3,2], cnt(3)=1, cnt(2)=1, 1 is evicted
      
Step 5: cache.get(1): return -1 (not found)
      
Step 6: cache.get(2): return 20, cache=[3,2], cnt(3)=1, cnt(2)=2
    
  

Constraints

    
0 ≤ capacity ≤ 10^4
    
0 ≤ key ≤ 10^5
    
0 ≤ value ≤ 10^9
    
At most 2 * 10^5 calls will be made to get and put.
    
Time Complexity: O(1) for all operations
    
Space Complexity: O(capacity)