Approach Summary
HashMap + doubly-linked list. Map stores key → node; list keeps usage order (head = MRU, tail = LRU). O(1) get and put.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n^2) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | get(int key) is O(1) | O(capacity) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(2^n)
Space: O(n^2)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class LRUCache {
private class Node{
int key;
int value;
Node next;
Node prev;
Node(int k, int v){
key = k;
value = v;
next = null;
prev = null;
}
}
private Map<Integer, Node> cache;
private int capacity;
private Node head;
private Node tail;
public LRUCache(int cap) {
capacity = cap;
cache = new HashMap<>();
head = new Node(-1, -1);
tail = new Node(-1, -1);
head.next = tail;
tail.prev = head;
}
public void addToFront(Node node){
node.next = head.next;
node.prev = head;
head.next = node;
node.next.prev = node;
}
private void removeNode(Node node){
node.prev.next = node.next;
node.next.prev = node.prev;
}
public int get(int key) {
if(cache.containsKey(key)){
Node node = cache.get(key);
removeNode(node);
addToFront(node);
return node.value;
}
return -1;
}
public void put(int key, int value) {
if(cache.containsKey(key)){
Node node = cache.get(key);
node.value = value;
removeNode(node);
addToFront(node);
}
else{
if(cache.size() >= capacity){
Node nodeToRemove = tail.prev;
removeNode(nodeToRemove);
cache.remove(nodeToRemove.key);
}
Node newNode = new Node(key, value);
addToFront(newNode);
cache.put(key, newNode);
}
}
}class LRUCache {
private:
struct Node{
int key;
int value;
Node* next;
Node* prev;
Node(int k, int v){
key = k;
value = v;
next = nullptr;
prev = nullptr;
}
};
unordered_map<int, Node*> cache;
int capacity;
Node* head;
Node* tail;
void addToFront(Node* node){
node->next = head->next;
node->prev = head;
head->next = node;
node->next->prev = node;
}
void removeNode(Node* node){
node->prev->next = node->next;
node->next->prev = node->prev;
}
public:
LRUCache(int cap) {
capacity = cap;
head = new Node(-1, -1);
tail = new Node(-1, -1);
head->next = tail;
tail->prev = head;
}
int get(int key) {
if(cache.find(key) != cache.end()){
Node* node = cache[key];
removeNode(node);
addToFront(node);
return node->value;
}
return -1;
}
void put(int key, int value) {
if(cache.find(key) != cache.end()){
Node* node = cache[key];
node->value = value;
removeNode(node);
addToFront(node);
}
else{
if(cache.size() >= capacity){
Node* nodeToRemove = tail->prev;
removeNode(nodeToRemove);
cache.erase(nodeToRemove->key);
delete nodeToRemove;
}
Node* newNode = new Node(key, value);
addToFront(newNode);
cache[key] = newNode;
}
}
};class Solution:
def lru_cache(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for LRU Cache
*/
function lruCache(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: get(int key) is O(1)
Space: O(capacity)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class LRUCache {
private class Node{
int key;
int value;
Node next;
Node prev;
Node(int k, int v){
key = k;
value = v;
next = null;
prev = null;
}
}
private Map<Integer, Node> cache;
private int capacity;
private Node head;
private Node tail;
public LRUCache(int cap) {
capacity = cap;
cache = new HashMap<>();
head = new Node(-1, -1);
tail = new Node(-1, -1);
head.next = tail;
tail.prev = head;
}
public void addToFront(Node node){
node.next = head.next;
node.prev = head;
head.next = node;
node.next.prev = node;
}
private void removeNode(Node node){
node.prev.next = node.next;
node.next.prev = node.prev;
}
public int get(int key) {
if(cache.containsKey(key)){
Node node = cache.get(key);
removeNode(node);
addToFront(node);
return node.value;
}
return -1;
}
public void put(int key, int value) {
if(cache.containsKey(key)){
Node node = cache.get(key);
node.value = value;
removeNode(node);
addToFront(node);
}
else{
if(cache.size() >= capacity){
Node nodeToRemove = tail.prev;
removeNode(nodeToRemove);
cache.remove(nodeToRemove.key);
}
Node newNode = new Node(key, value);
addToFront(newNode);
cache.put(key, newNode);
}
}
}class LRUCache {
private:
struct Node{
int key;
int value;
Node* next;
Node* prev;
Node(int k, int v){
key = k;
value = v;
next = nullptr;
prev = nullptr;
}
};
unordered_map<int, Node*> cache;
int capacity;
Node* head;
Node* tail;
void addToFront(Node* node){
node->next = head->next;
node->prev = head;
head->next = node;
node->next->prev = node;
}
void removeNode(Node* node){
node->prev->next = node->next;
node->next->prev = node->prev;
}
public:
LRUCache(int cap) {
capacity = cap;
head = new Node(-1, -1);
tail = new Node(-1, -1);
head->next = tail;
tail->prev = head;
}
int get(int key) {
if(cache.find(key) != cache.end()){
Node* node = cache[key];
removeNode(node);
addToFront(node);
return node->value;
}
return -1;
}
void put(int key, int value) {
if(cache.find(key) != cache.end()){
Node* node = cache[key];
node->value = value;
removeNode(node);
addToFront(node);
}
else{
if(cache.size() >= capacity){
Node* nodeToRemove = tail->prev;
removeNode(nodeToRemove);
cache.erase(nodeToRemove->key);
delete nodeToRemove;
}
Node* newNode = new Node(key, value);
addToFront(newNode);
cache[key] = newNode;
}
}
};class Solution:
def lru_cache(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for LRU Cache
*/
function lruCache(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Hash Map / Set Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Hash Map / Set pattern.
Framework: Hash Map / Set • Interview Prep: Tested at FAANG •
Explore all Hash Map / Set problems & invariants →
How to Recognize This Pattern
- "Design a cache with O(1) get/put and eviction"
- Ordered hash map or manual DLL + map
Complexity Analysis
Time Complexity
O(1)
Space Complexity
O(capacity)
Tags
Hash Table Linked List Design
Frequently Asked Questions
What is the optimal time and space complexity for LRU Cache?
The optimal solution for LRU Cache achieves O(1) runtime complexity and requires O(capacity) auxiliary space. Time Complexity is get(int key) is O(1) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(capacity) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that LRU Cache requires the Hash Map / Set pattern?
In technical interviews, look for key problem signals: "Design a cache with O(1) get/put and eviction"; Ordered hash map or manual DLL + map. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Hash Map / Set strategy.
What critical edge cases must be handled in LRU Cache?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask LRU Cache?
LRU Cache is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Microsoft, Google, Facebook, Apple, Uber, Stripe.