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Problem solution · Java

LFU Cache

LFU Cache: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
56 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For LFU Cache, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 56 lines of Java from the credited upstream file 460.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeLFU Cache · JavaJava
Use this to learn the idea, then write your own version.
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 an LRU key from `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);  }}

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