Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

  • LRUCache(int capacity) Initialize the LRU cache with positive size capacity.
  • int get(int key) Return the value of the key if the key exists, otherwise return -1.
  • void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key.

The functions get and put must each run in O(1) average time complexity.

Examples

Example 1:

Input
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, null, -1, 3, 4]

Explanation
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // cache is {1=1}
lRUCache.put(2, 2); // cache is {1=1, 2=2}
lRUCache.get(1);    // return 1
lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2);    // returns -1 (not found)
lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(1);    // return -1 (not found)
lRUCache.get(3);    // return 3
lRUCache.get(4);    // return 4

Constraints

  • 1 <= capacity <= 3000
  • 0 <= key <= 10^4
  • 0 <= value <= 10^5
  • At most 2 * 10^5 calls will be made to get and put.

Thinking Process

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

  • Draw pointers before rewriting links.
  • Dummy head simplifies insert/delete at the head.
  • Slow/fast pointers find middle or detect cycles in one pass.
Linked list: pointer walk 1 2 3 slow → → fast (2x speed)

Common Approaches

Typical techniques for this pattern:

Approach Time Space Notes
Iterative pointer walk (this problem) O(n) O(1) Traversal, insertion
Dummy head node O(n) O(1) Simplify head-edge cases
Reversal (3-pointer) O(n) O(1) Reverse sublist or full list
Slow/fast pointers O(n) O(1) Middle, cycle, merge lists

Solution

Time Complexity: O(1) for both get and put
Space Complexity: O(capacity)

We use a combination of hash map and doubly linked list to achieve O(1) operations. The hash map stores key-to-node mappings, and the doubly linked list maintains the order of recently used items.

Solution 1: Using list (Recommended - C++20 Optimized)

using namespace std;

class LRUCache {
private:
    int capacity_;
    unordered_map<int, list<pair<int, int>>::iterator> cache_;
    list<pair<int, int>> lru_list_;

    // Helper to move node to front (most recently used)
    void moveToFront(list<pair<int, int>>::iterator it) {
        if (it != lru_list_.begin()) {
            lru_list_.splice(lru_list_.begin(), lru_list_, it);
        }
    }

public:
    explicit LRUCache(int capacity) 
        : capacity_(capacity) 
    {
        cache_.reserve(capacity_);  // Pre-allocate hash map
    }

    int get(int key) {
        auto it = cache_.find(key);
        if (it == cache_.end()) {
            return -1;
        }

        // Move to front (most recently used)
        moveToFront(it->second);
        return it->second->second;
    }

    void put(int key, int value) {
        auto it = cache_.find(key);
        
        if (it != cache_.end()) {
            // Update existing key
            it->second->second = value;
            moveToFront(it->second);
        } else {
            // Add new key
            if (cache_.size() >= capacity_) {
                // Evict least recently used (back of list)
                auto [lru_key, _] = lru_list_.back();
                cache_.erase(lru_key);
                lru_list_.pop_back();
            }
            
            // Insert at front
            lru_list_.emplace_front(key, value);
            cache_[key] = lru_list_.begin();
        }
    }
};

Solution Explanation

Approach: Iterative pointer walk (this problem)

Key idea: Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

How the code works:

  • Draw pointers before rewriting links.
  • Dummy head simplifies insert/delete at the head.
  • Slow/fast pointers find middle or detect cycles in one pass.
Operation Time Space
get(key) O(1) O(1)
put(key, value) O(1) O(1)
Overall O(1) O(capacity)

Solution 2: Custom Doubly Linked List (C++20 Optimized)

using namespace std;

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) {}
    };

    int capacity_;
    unordered_map<int, Node*> cache_;
    
    // Dummy head and tail for easier list manipulation
    unique_ptr<Node> head_;
    unique_ptr<Node> tail_;

    // Add node right before tail (most recently used)
    void addNode(Node* node) {
        Node* prev_end = tail_->prev;
        prev_end->next = node;
        node->prev = prev_end;
        node->next = tail_.get();
        tail_->prev = node;
    }

    // Remove node from list
    void removeNode(Node* node) {
        node->prev->next = node->next;
        node->next->prev = node->prev;
    }

    // Move node to end (most recently used)
    void moveToEnd(Node* node) {
        removeNode(node);
        addNode(node);
    }

public:
    explicit LRUCache(int capacity) 
        : capacity_(capacity)
        , head_(make_unique<Node>(-1, -1))
        , tail_(make_unique<Node>(-1, -1))
    {
        head_->next = tail_.get();
        tail_->prev = head_.get();
        cache_.reserve(capacity_);
    }

    ~LRUCache() {
        // Clean up nodes
        Node* current = head_->next;
        while (current != tail_.get()) {
            Node* next = current->next;
            delete current;
            current = next;
        }
    }

    // Delete copy constructor and assignment
    LRUCache(const LRUCache&) = delete;
    LRUCache& operator=(const LRUCache&) = delete;

    int get(int key) {
        auto it = cache_.find(key);
        if (it == cache_.end()) {
            return -1;
        }

        Node* node = it->second;
        moveToEnd(node);
        return node->value;
    }

    void put(int key, int value) {
        auto it = cache_.find(key);
        
        if (it != cache_.end()) {
            // Update existing
            Node* node = it->second;
            node->value = value;
            moveToEnd(node);
        } else {
            // Add new
            if (cache_.size() >= capacity_) {
                // Evict least recently used (head->next)
                Node* lru = head_->next;
                removeNode(lru);
                cache_.erase(lru->key);
                delete lru;
            }
            
            Node* newNode = new Node(key, value);
            addNode(newNode);
            cache_[key] = newNode;
        }
    }
};

Solution 3: Most Optimized with Move Semantics

#include <unordered_map>
#include <list>
#include <utility>

class LRUCache {
private:
    int capacity_;
    std::unordered_map<int, std::list<std::pair<int, int>>::iterator> cache_;
    std::list<std::pair<int, int>> lru_list_;

public:
    explicit LRUCache(int capacity) 
        : capacity_(capacity) 
    {
        cache_.reserve(capacity_);
    }

    [[nodiscard]] int get(int key) {
        const auto it = cache_.find(key);
        if (it == cache_.end()) {
            return -1;
        }

        // Move to front using splice (O(1))
        lru_list_.splice(lru_list_.begin(), lru_list_, it->second);
        return it->second->second;
    }

    void put(int key, int value) {
        auto it = cache_.find(key);
        
        if (it != cache_.end()) {
            // Update and move to front
            it->second->second = value;
            lru_list_.splice(lru_list_.begin(), lru_list_, it->second);
        } else {
            // Check capacity
            if (cache_.size() >= capacity_) {
                // Evict LRU (back of list)
                cache_.erase(lru_list_.back().first);
                lru_list_.pop_back();
            }
            
            // Insert at front
            lru_list_.emplace_front(key, value);
            cache_[key] = lru_list_.begin();
        }
    }
};

Key Optimizations (C++20)

  1. list::splice(): O(1) operation to move nodes without copying
  2. unordered_map::reserve(): Pre-allocates hash map to avoid rehashing
  3. explicit constructor: Prevents implicit conversions
  4. Structured bindings: Cleaner code with auto [key, value]
  5. emplace_front(): Constructs in-place, avoiding copies
  6. Move semantics: Efficient transfer of ownership

Data Structure Design

Hash Map:          Doubly Linked List:
key -> iterator    [head] <-> [1,1] <-> [2,2] <-> [tail]
                   (LRU)                (MRU)

Operation Flow

Get Operation:

  1. Look up key in hash map → O(1)
  2. If found, move node to front (most recently used) → O(1)
  3. Return value

Put Operation:

  1. Look up key in hash map → O(1)
  2. If exists: update value and move to front → O(1)
  3. If new:
    • Check capacity
    • If full: remove back node (LRU) → O(1)
    • Insert at front → O(1)

Example Walkthrough

capacity = 2

put(1, 1):  cache = {1: [1,1]}
            list: [head] <-> [1,1] <-> [tail]

put(2, 2):  cache = {1: [1,1], 2: [2,2]}
            list: [head] <-> [1,1] <-> [2,2] <-> [tail]

get(1):     Move [1,1] to front
            list: [head] <-> [2,2] <-> [1,1] <-> [tail]
            return 1

put(3, 3):  Evict [2,2] (LRU), add [3,3] at front
            cache = {1: [1,1], 3: [3,3]}
            list: [head] <-> [3,3] <-> [1,1] <-> [tail]

Why std::list is Preferred

  1. splice() is O(1): Moves nodes without copying
  2. Automatic memory management: No manual node deletion
  3. Less error-prone: No pointer management
  4. Better cache locality: Standard library optimizations
  5. Cleaner code: Less boilerplate

Common Mistakes

  1. Capacity = 1: Only one item can exist
  2. Get non-existent key: Returns -1
  3. Update existing key: Moves to front, doesn’t increase size

  4. Multiple puts: Evicts oldest when capacity exceeded

  5. Not moving to front on get: Must update access order
  6. Wrong eviction order: Remove from back (LRU), not front
  7. Memory leaks: Forgetting to delete nodes in custom implementation
  8. Not updating iterator: After list modification, iterators may be invalid
  9. Copying instead of moving: Use splice() or move semantics

Key Takeaways

  • Pattern: Iterative pointer walk (this problem)
  • Draw pointers before rewriting links.
  • Dummy head simplifies insert/delete at the head.

References