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-iterator mappings, and the doubly linked list maintains the order of recently used items.

Solution: Using std::list with splice

class LRUCache {
    private final int cap;
    private final Map<Integer, Integer> map = new LinkedHashMap<>(16, 0.75f, true);

    public LRUCache(int capacity) {
        cap = capacity;
    }

    public int get(int key) {
        return map.getOrDefault(key, -1);
    }

    public void put(int key, int value) {
        if (map.containsKey(key)) {
            map.remove(key);
        } else if (map.size() == cap) {
            int eldest = map.keySet().iterator().next();
            map.remove(eldest);
        }
        map.put(key, value);
    }
}```

### 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)** |
## Thread-Safe LRU Cache

Thread-safe version using mutex for concurrent access.

```java
class LRUCache {
    class Node {
        int key, value;
        Node prev, next;
        Node(int k, int v) { key = k; value = v; }
    }

    private final int cap;
    private final Map<Integer, Node> map = new HashMap<>();
    private final Node head = new Node(0, 0), tail = new Node(0, 0);

    public LRUCache(int capacity) {
        cap = capacity;
        head.next = tail;
        tail.prev = head;
    }

    public int get(int key) {
        if (!map.containsKey(key)) return -1;
        Node node = map.get(key);
        moveToEnd(node);
        return node.value;
    }

    public void put(int key, int value) {
        if (map.containsKey(key)) {
            Node node = map.get(key);
            node.value = value;
            moveToEnd(node);
            return;
        }
        if (map.size() == cap) {
            Node lru = head.next;
            remove(lru);
            map.remove(lru.key);
        }
        Node node = new Node(key, value);
        map.put(key, node);
        insertEnd(node);
    }

    private void remove(Node node) {
        node.prev.next = node.next;
        node.next.prev = node.prev;
    }

    private void insertEnd(Node node) {
        node.prev = tail.prev;
        node.next = tail;
        tail.prev.next = node;
        tail.prev = node;
    }

    private void moveToEnd(Node node) {
        remove(node);
        insertEnd(node);
    }
}```

### Thread-Safe Implementation Details

1. **`shared_mutex`**: Allows multiple concurrent reads when no writes are happening
2. **`unique_lock<shared_mutex>`**: Exclusive lock for both `get()` and `put()` since they modify the list
3. **`shared_lock<shared_mutex>`**: Shared lock for read-only operations like `size()`
4. **Note**: The thread-safe `put()` calls `get()` which requires careful lock management - both methods use exclusive locks

### Solution Explanation
### Data Structure Design

Hash Map: Doubly Linked List: key -> (value, iterator) [1] <-> [2] <-> [3] (LRU) (MRU)


### Operation Flow

**Get Operation:**
1. Look up key in hash map → O(1)
2. If found, move node to end (most recently used) using `splice()` → O(1)
3. Return value

**Put Operation:**
1. Check if key exists via `get()` → O(1)
2. If exists: update value (already moved to end by `get()`) → O(1)
3. If new:
   - Check capacity
   - If full: remove front node (LRU) → O(1)
   - Insert at end → O(1)

### Example Walkthrough

capacity = 2

put(1, 1): cache = {1: (1, it1)} list: [1]

put(2, 2): cache = {1: (1, it1), 2: (2, it2)} list: [1] <-> [2]

get(1): Move 1 to end using splice list: [2] <-> [1] return 1

put(3, 3): get(3) returns -1, capacity full Remove front (key 2), insert 3 at end cache = {1: (1, it1), 3: (3, it3)} list: [1] <-> [3] ```

Complexity

| Operation | Time | Space | |———–|——|——-| | get(key) | O(1) | O(1) | | put(key, value) | O(1) | O(1) | | Overall | O(1) | O(capacity) |

Why This Approach Works

  1. Hash map: Provides O(1) lookup by key
  2. Doubly linked list: Maintains insertion order and allows O(1) removal/insertion
  3. Iterator storage: Hash map stores iterators to list nodes for O(1) access
  4. splice() optimization: Moves nodes without copying, maintaining O(1) complexity

Common Mistakes

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

  4. Multiple puts: Evicts oldest when capacity exceeded

  5. Not moving to end on get: Must update access order
  6. Wrong eviction order: Remove from front (LRU), not back
  7. Invalid iterator after modification: Use splice() which preserves iterators
  8. Not handling existing key in put: Must check and update, not just insert

Key Takeaways

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

References