Implement an LFU cache
Design and implement a data structure for a 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.
class LFUCache {
public LFUCache(int capacity) {
this.capacity = capacity;
}
public int get(int key) {
if (!keyToVal.containsKey(key))
return -1;
final int freq = keyToFreq.get(key);
freqToLRUKeys.get(freq).remove(key);
if (freq == minFreq && freqToLRUKeys.get(freq).isEmpty()) {
freqToLRUKeys.remove(freq);
++minFreq;
}
// Increase key's freq by 1
// Add this key to next freq's list
putFreq(key, freq + 1);
return keyToVal.get(key);
}
public void put(int key, int value) {
if (capacity == 0)
return;
if (keyToVal.containsKey(key)) {
keyToVal.put(key, value);
get(key); // Update key's count
return;
}
if (keyToVal.size() == capacity) {
// Evict LRU key from the minFreq list
final int keyToEvict = freqToLRUKeys.get(minFreq).iterator().next();
freqToLRUKeys.get(minFreq).remove(keyToEvict);
keyToVal.remove(keyToEvict);
}
minFreq = 1;
putFreq(key, minFreq); // Add new key and freq
keyToVal.put(key, value); // Add new key and value
}
private int capacity;
private int minFreq = 0;
private Map<Integer, Integer> keyToVal = new HashMap<>();
private Map<Integer, Integer> keyToFreq = new HashMap<>();
private Map<Integer, LinkedHashSet<Integer>> freqToLRUKeys = new HashMap<>();
private void putFreq(int key, int freq) {
keyToFreq.put(key, freq);
freqToLRUKeys.putIfAbsent(freq, new LinkedHashSet<>());
freqToLRUKeys.get(freq).add(key);
}
}Related Problems
Given an integer array nums, return an integer array counts where counts[i] is the number of smaller elements to the right of nums[i].