Here, We see LFU Cache LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
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LFU Cache LeetCode Solution
Table of Contents
Problem Statement
Design and implement a data structure for a Least Frequently Used (LFU) cache.
Implement the LFUCache class:
LFUCache(int capacity)Initializes the object with thecapacityof the data structure.int get(int key)Gets the value of thekeyif thekeyexists in the cache. Otherwise, returns-1.void put(int key, int value)Update the value of thekeyif present, or inserts thekeyif not already present. When the cache reaches itscapacity, 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 usedkeywould 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
LFU Cache LeetCode Solution C++
class LFUCache {
int capacity, minFreq;
struct Element {
int value, fr;
list<int>::iterator it;
};
unordered_map<int, Element> cache;
unordered_map<int, list<int> > freq;
public:
LFUCache(int capacity) {
this->capacity = capacity;
}
int get(int key) {
if (!cache.count(key)) return -1;
freq[cache[key].fr].erase(cache[key].it);
cache[key].fr++;
freq[cache[key].fr].push_back(key);
cache[key].it = prev(freq[ cache[key].fr].end());
minFreq += freq[minFreq].empty();
return cache[key].value;
}
void put(int key, int value) {
if (!capacity) return;
if (get(key) != -1) {
cache[key].value = value;
return;
}
if (cache.size() == capacity){
cache.erase(freq[minFreq].front());
freq[minFreq].pop_front();
}
freq[1].push_back(key);
cache[key] = {value, 1, prev(freq[1].end())};
minFreq = 1;
}
};Code language: JavaScript (javascript)LFU Cache LeetCode Solution Java
class LFUCache {
private final Map<Integer, CachedElement<Integer, Integer>> elements;
private final TreeMap<Integer, LinkedList<Integer>> frequencies = new TreeMap<>();
private final int capacity;
public LFUCache(int capacity) {
this.capacity = capacity;
this.elements = new HashMap<>(capacity);
}
public int get(int key) {
CachedElement<Integer, Integer> element = this.elements.get(key);
if (element == null) {
return -1;
}
incrementFrequency(element);
return element.value;
}
public void put(int key, int value) {
if (!this.elements.containsKey(key) && this.capacity == this.elements.size()) {
this.removeLFU();
}
CachedElement<Integer, Integer> element = this.elements.get(key);
if (element == null) {
element = new CachedElement<>(key, value);
this.addNew(element);
} else {
element.value = value;
incrementFrequency(element);
}
this.elements.put(key, element);
}
private void addNew(CachedElement<Integer, Integer> element) {
LinkedList<Integer> elementsOfOneFrequency = this.frequencies.getOrDefault(
element.accessCount,
new LinkedList<>()
);
elementsOfOneFrequency.addFirst(element.key);
this.frequencies.put(element.accessCount, elementsOfOneFrequency);
}
private void removeLFU() {
int elementsOfMinFrequencyKey = frequencies.firstKey();
LinkedList<Integer> elementsOfMinFrequency = frequencies.remove(elementsOfMinFrequencyKey);
int lfuKey = elementsOfMinFrequency.removeLast();
if (!elementsOfMinFrequency.isEmpty()) {
this.frequencies.put(elementsOfMinFrequencyKey, elementsOfMinFrequency);
}
this.elements.remove(lfuKey);
}
private void incrementFrequency(CachedElement<Integer, Integer> element) {
LinkedList<Integer> elementsOfCurrentFrequency = this.frequencies.remove(element.accessCount);
elementsOfCurrentFrequency.remove(element.key);
if (!elementsOfCurrentFrequency.isEmpty()) {
this.frequencies.put(element.accessCount, elementsOfCurrentFrequency);
}
element.accessCount++;
LinkedList<Integer> elementsOfHigherFrequency = this.frequencies.getOrDefault(
element.accessCount,
new LinkedList<>()
);
elementsOfHigherFrequency.addFirst(element.key);
this.frequencies.put(element.accessCount, elementsOfHigherFrequency);
}
private static class CachedElement<K, V> {
private final K key;
private V value;
private int accessCount;
CachedElement(K key, V value) {
this.key = key;
this.value = value;
this.accessCount = 1;
}
}
}Code language: JavaScript (javascript)LFU Cache LeetCode Solution JavaScript
var LFUCache = function (capacity) {
this.MIN_FREQUENCY = 1;
this.currentMinFrequency = Number.MAX_VALUE;
this.capacity = capacity;
this.savedCapacity = capacity; // to check case with capacity === 0
this.itemsByFrequency = new Map();
this.frequencyByKey = new Map();
};
LFUCache.prototype.getValue = function (key) {
if (!this.frequencyByKey.has(key)) return -1;
const frequencyKey = this.frequencyByKey.get(key);
return this.itemsByFrequency.get(frequencyKey).get(key);
}
LFUCache.prototype.addItem = function (key, value) {
this.currentMinFrequency = this.MIN_FREQUENCY;
this.frequencyByKey.set(key, this.MIN_FREQUENCY);
const items = this.itemsByFrequency.get(this.MIN_FREQUENCY) || new Map();
items.set(key, value);
this.itemsByFrequency.set(this.MIN_FREQUENCY, items);
}
LFUCache.prototype.updateItem = function (key, value) {
if (value === -1) return -1;
let currentItemFreaquency = this.frequencyByKey.get(key);
const currentItems = this.itemsByFrequency.get(currentItemFreaquency);
currentItems.delete(key);
if (!currentItems.size && this.currentMinFrequency === currentItemFreaquency) this.currentMinFrequency++;
if (!currentItems.size) this.itemsByFrequency.delete(currentItemFreaquency);
if (currentItems.size) this.itemsByFrequency.set(currentItemFreaquency, currentItems);
this.frequencyByKey.delete(key);
const updatedItemFrequency = currentItemFreaquency + 1;
this.frequencyByKey.set(key, updatedItemFrequency);
const updatedItems = this.itemsByFrequency.get(updatedItemFrequency) || new Map();
updatedItems.set(key, value);
this.itemsByFrequency.set(updatedItemFrequency, updatedItems);
}
LFUCache.prototype.removeLFUItem = function () {
if (!this.itemsByFrequency.size || !this.frequencyByKey.size) return null;
const LFUItems = this.itemsByFrequency.get(this.currentMinFrequency);
const LFUKey = LFUItems.keys().next().value;
this.frequencyByKey.delete(Number(LFUKey));
if (LFUItems.size === 1) {
this.itemsByFrequency.delete(this.currentMinFrequency);
this.currentMinFrequency = this.itemsByFrequency.keys().next().value;
} else {
LFUItems.delete(LFUKey);
this.itemsByFrequency.set(this.currentMinFrequency, LFUItems);
}
}
LFUCache.prototype.put = function (key, value) {
if (this.frequencyByKey.has(key)) {
this.updateItem(key, value);
return null;
}
if (this.capacity) {
this.addItem(key, value);
this.capacity--;
return null;
}
if (!this.capacity && this.savedCapacity) {
this.removeLFUItem(key, value);
this.addItem(key, value);
return null;
}
};
LFUCache.prototype.get = function (key) {
this.updateItem(key, this.getValue(key));
return this.getValue(key);
};Code language: JavaScript (javascript)LFU Cache LeetCode Solution Python
class ListNode(object):
def __init__(self, key, value):
self.key = key
self.val = value
self.freq = 1
class LFUCache(object):
def __init__(self, capacity):
self.capacity = capacity
self.cache = dict()
self.usage = collections.defaultdict(collections.OrderedDict)
self.LF = 0
def get(self, key):
if key not in self.cache:return -1
node = self.cache[key]
self.update(node, node.val)
return node.val
def put(self, key, value):
if self.capacity == 0: return
if key not in self.cache:
if len(self.cache) >= self.capacity:
k, v = self.usage[self.LF].popitem(last=False)
self.cache.pop(k)
node = ListNode(key, value)
self.cache[key] = node
self.usage[1][key] = value
self.LF = 1
else:
node = self.cache[key]
node.val = value
self.update(node, value)
def update(self, node, newVal):
k, f = node.key, node.freq
self.usage[f].pop(k)
if not self.usage[f] and self.LF == f:
self.LF += 1
self.usage[f+1][k] = newVal
node.freq += 1