460. Leetcode Solution in CPP
Tags: cpp
Published: 2023-01-29 03:40:18

==============================================================

Description:
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.
 ==============================================================
Solution:
```cpp
struct Node {
  int key;
  int value;
  int freq;
  list<int>::const_iterator it;
};

class LFUCache {
 public:
  LFUCache(int capacity) : capacity(capacity), minFreq(0) {}

  int get(int key) {
    if (!keyToNode.count(key))
      return -1;

    Node& node = keyToNode[key];
    touch(node);
    return node.value;
  }

  void put(int key, int value) {
    if (capacity == 0)
      return;
    if (keyToNode.count(key)) {
      Node& node = keyToNode[key];
      node.value = value;
      touch(node);
      return;
    }

    if (keyToNode.size() == capacity) {
      // evict LRU key from the minFreq list
      const int keyToEvict = freqToList[minFreq].back();
      freqToList[minFreq].pop_back();
      keyToNode.erase(keyToEvict);
    }

    minFreq = 1;
    freqToList[1].push_front(key);
    keyToNode[key] = {key, value, 1, cbegin(freqToList[1])};
  }

 private:
  int capacity;
  int minFreq;
  unordered_map<int, Node> keyToNode;
  unordered_map<int, list<int>> freqToList;

  void touch(Node& node) {
    // update the node's frequency
    const int prevFreq = node.freq;
    const int newFreq = ++node.freq;

    // remove the iterator from prevFreq's list
    freqToList[prevFreq].erase(node.it);
    if (freqToList[prevFreq].empty()) {
      freqToList.erase(prevFreq);
      // update minFreq if needed
      if (prevFreq == minFreq)
        ++minFreq;
    }

    // insert the key to the front of newFreq's list
    freqToList[newFreq].push_front(node.key);
    node.it = cbegin(freqToList[newFreq]);
  }
};


/**
 * Your LFUCache object will be instantiated and called as such:
 * LFUCache* obj = new LFUCache(capacity);
 * int param_1 = obj->get(key);
 * obj->put(key,value);
 */
```
=================================================================

#leetcode
#challenge
Here is the link for the problem:
https://leetcode.com/problems/lfu-cache/