In this guide, you will get 460. LFU Cache LeetCode Solution with the best time and space complexity. The solution to LFU Cache problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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.
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:
1 <= capacity <= 104
0 <= key <= 105
0 <= value <= 109
At most 2 * 105 calls will be made to get and put.
Complexity Analysis
Time Complexity: O(1)
Space Complexity: O(n)
460. LFU Cache LeetCode Solution in C++
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) {
const auto it = keyToNode.find(key);
if (it == keyToNode.cend())
return -1;
Node& node = it->second;
touch(node);
return node.value;
}
void put(int key, int value) {
if (capacity == 0)
return;
if (const auto it = keyToNode.find(key); it != keyToNode.cend()) {
Node& node = it->second;
node.value = value;
touch(node);
return;
}
if (keyToNode.size() == capacity) {
// Evict an LRU key from `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, freqToList[1].cbegin()};
}
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 = freqToList[newFreq].cbegin();
}
}; /* code provided by PROGIEZ */
460. LFU Cache LeetCode Solution in Java
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);
}
} // code provided by PROGIEZ
