Least Frequently Used Cache - Amazon Top Interview Questions
Problem Statement :
Implement a least frequently used cache with the following methods: LFUCache(int capacity) constructs a new instance of a LFU cache with the given capacity. get(int key) retrieves the value associated with the key key. If it does not exist, return -1. As a side effect, this key's usage is incremented (used for eviction). set(int key, int val) updates the key key with value val. If updating this key-value pair exceeds capacity, then evicts the least frequently used key-value pair. If there is more than one key that's been used least frequently used, it should use the least recently used key. Each method should be done in \mathcal{O}(1)O(1) time. Constraints n ≤ 100,000 where n is the number of calls to get and set. Example 1 Input methods = ["constructor", "set", "get", "set", "set", "set", "get", "get"] arguments = [[3], [1, 10], [1], [2, 20], [3, 30], [4, 40], [3], [2]]` Output [None, None, 10, None, None, None, 30, -1] Explanation We create an LFUCache of capacity 3. Set key of 1 to value 10. Size of cache is now 1 We get 1 which has value of 10 Set key of 2 to value 20. Size of cache is now 2 Set key of 3 to value 30. Size of cache is now 3 Set key of 4 to value 40. Size exceeds capacity, so now we evict the least frequently used key. Since both 2 and 3 are tied for being used the least, we evict the least recently used key which is 2. We get 3 which has value of 30 We get 2 which has been evicted so we return -1
Solution :
Solution in C++ :
void set(int key, int value) {
// Scenario 1:
if (!keyMap.count(key)) {
// Case 1:
if (keyMap.size() < cap) {
freqMap[1].emplace_back(key, value);
keyMap[key] = {1, --freqMap[1].end()};
lowestFreq = 1;
}
// Case 2:
if (keyMap.size() == cap) {
if (lowestFreq == 0) return;
auto [k, v] = freqMap[lowestFreq].front();
freqMap[lowestFreq].erase(freqMap[lowestFreq].begin());
keyMap.erase(k);
set(key, value);
}
}
// Scenario 2:
if(keyMap.count(key) {
auto [freq, it] = keyMap[key];
if (freqMap[freq].size() == 1 and freq == lowestFreq) lowestFreq++;
freqMap[freq].erase(it);
freqMap[freq+1].emplace_back(key, value);
keyMap[key] = {freq+1, --freqMap[freq+1].end()};
}
}
Solution in Java :
import java.util.*;
class LFUCache {
private Map<Integer, Integer> values;
private Map<Integer, Integer> freq;
private Map<Integer, LinkedHashSet<Integer>> lists;
private int capacity;
private int minIndex = -1; // to keep track of lists (index)
public LFUCache(int capacity) {
this.capacity = capacity;
values = new HashMap<>();
freq = new HashMap<>();
lists = new HashMap<>();
lists.put(1, new LinkedHashSet<>()); // create 1 count bucket
}
public int get(int key) {
// key not found
if (!values.containsKey(key))
return -1;
// update frequency counts
int count = freq.get(key);
freq.put(key, count + 1);
// remove from <count> bucket lists
lists.get(count).remove(key);
// update min to pick next bucket from list
if (count == minIndex && lists.get(count).size() == 0)
minIndex++;
// add it new count bucket lists
lists.computeIfAbsent(count + 1, f -> new LinkedHashSet());
lists.get(count + 1).add(key);
return values.get(key);
}
public void set(int key, int val) {
// not enough capacity
if (capacity <= 0)
return;
// Case 1: if key exists
if (values.containsKey(key)) {
values.put(key, val);
get(key); // update count
return;
}
// case 2: if capacity is Full
// evict policy
if (values.size() >= capacity) {
int evit = lists.get(minIndex).iterator().next();
lists.get(minIndex).remove(evit);
values.remove(evit);
}
values.put(key, val);
freq.put(key, 1);
minIndex = 1; // reset min index
lists.get(minIndex).add(key);
}
}
Solution in Python :
from collections import defaultdict, OrderedDict
class Node:
def __init__(self, key: int, value: int, count: int = 1):
self.key = key
self.value = value
self.count = count
class LFUCache:
def __init__(self, capacity: int):
self._size = 0
self._capacity = capacity
self._nodes = {}
self._counts = defaultdict(OrderedDict)
self._min_count = 0
def _touch(self, node: Node):
self._counts[node.count].popitem(node.key)
if not self._counts[self._min_count]:
self._min_count += 1
node.count += 1
self._counts[node.count][node.key] = node
def get(self, key: int) -> int:
if key not in self._nodes:
return -1
node = self._nodes[key]
self._touch(node)
return node.value
def set(self, key: int, value: int):
if not self._capacity:
return
if key in self._nodes:
node = self._nodes[key]
self._touch(node)
node.value = value
else:
if self._size == self._capacity:
evict_key, _ = self._counts[self._min_count].popitem(last=False)
self._nodes.pop(evict_key)
self._size -= 1
node = Node(key, value)
self._nodes[key] = node
self._counts[1][key] = node
self._min_count = 1
self._size += 1
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