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.


n ≤ 100,000 where n is the number of calls to get and set.

Example 1


methods = ["constructor", "set", "get", "set", "set", "set", "get", "get"]

arguments = [[3], [1, 10], [1], [2, 20], [3, 30], [4, 40], [3], [2]]`


[None, None, 10, None, None, None, 30, -1]


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();
            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+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
        // update min to pick next bucket from list
        if (count == minIndex && lists.get(count).size() == 0)

        // 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)

        // Case 1: if key exists
        if (values.containsKey(key)) {
            values.put(key, val);
            get(key); // update count

        // case 2: if capacity is Full
        // evict policy
        if (values.size() >= capacity) {
            int evit = lists.get(minIndex).iterator().next();

        values.put(key, val);
        freq.put(key, 1);
        minIndex = 1; // reset min index

                        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):
        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]
        return node.value

    def set(self, key: int, value: int):
        if not self._capacity:

        if key in self._nodes:
            node = self._nodes[key]
            node.value = value
            if self._size == self._capacity:
                evict_key, _ = self._counts[self._min_count].popitem(last=False)
                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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