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146. LRU Cache

mediumAsked at Snap

Snap Memories and story previews rely on an LRU layer to decide which media blobs to keep in RAM — this problem is the exact data-structure contract that cache layer implements.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Design a data structure that follows the Least Recently Used (LRU) cache constraints. Implement the LRUCache class with capacity, get(key) returning -1 if missing, and put(key, value) which evicts the least recently used key when at capacity. Both operations must run in O(1) average time.

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

Examples

Example 1

Input
LRUCache(2); put(1,1); put(2,2); get(1) → 1; put(3,3); get(2) → -1; put(4,4); get(1) → -1; get(3) → 3; get(4) → 4
Output
[1,-1,-1,3,4]

Example 2

Input
LRUCache(1); put(2,1); get(2) → 1; put(3,2); get(2) → -1; get(3) → 2
Output
[1,-1,2]

Approaches

1. Ordered Map (suboptimal)

Use a Map (insertion-ordered in JS) and re-insert on every get/put to track recency. Delete + re-insert is O(1) amortized in V8 but the approach is conceptually clear as a stepping stone.

Time
O(1) amortized
Space
O(capacity)
class LRUCache {
  constructor(capacity) {
    this.cap = capacity;
    this.map = new Map();
  }
  get(key) {
    if (!this.map.has(key)) return -1;
    const val = this.map.get(key);
    this.map.delete(key);
    this.map.set(key, val);
    return val;
  }
  put(key, value) {
    if (this.map.has(key)) this.map.delete(key);
    this.map.set(key, value);
    if (this.map.size > this.cap) {
      this.map.delete(this.map.keys().next().value);
    }
  }
}

Tradeoff:

2. HashMap + Doubly Linked List

HashMap gives O(1) lookup; doubly linked list (with dummy head/tail sentinels) gives O(1) move-to-front and O(1) evict-from-tail. This is the canonical interview answer Snap expects at L4+.

Time
O(1)
Space
O(capacity)
class LRUCache {
  constructor(capacity) {
    this.cap = capacity;
    this.map = new Map();
    this.head = { key: 0, val: 0, prev: null, next: null };
    this.tail = { key: 0, val: 0, prev: null, next: null };
    this.head.next = this.tail;
    this.tail.prev = this.head;
  }
  _remove(node) {
    node.prev.next = node.next;
    node.next.prev = node.prev;
  }
  _insertFront(node) {
    node.next = this.head.next;
    node.prev = this.head;
    this.head.next.prev = node;
    this.head.next = node;
  }
  get(key) {
    if (!this.map.has(key)) return -1;
    const node = this.map.get(key);
    this._remove(node);
    this._insertFront(node);
    return node.val;
  }
  put(key, value) {
    if (this.map.has(key)) this._remove(this.map.get(key));
    const node = { key, val: value, prev: null, next: null };
    this._insertFront(node);
    this.map.set(key, node);
    if (this.map.size > this.cap) {
      const lru = this.tail.prev;
      this._remove(lru);
      this.map.delete(lru.key);
    }
  }
}

Tradeoff:

Snap-specific tips

Snap's infrastructure team almost always follows up with 'how would you make this thread-safe?' and 'what eviction policy would you swap in for Snap Stories where recency matters less than access frequency?' Have answers for both.

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