22. LRU Cache
mediumAsked at AdyenDesign a data structure that supports get and put in O(1) with least-recently-used eviction.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Design a data structure that follows the constraints of a Least Recently Used (LRU) cache. Implement get(key) and put(key, value) such that both run in O(1) average time complexity.
Constraints
1 <= capacity <= 30000 <= key, value <= 10^4At most 2 * 10^5 calls.
Examples
Example 1
LRUCache(2); put(1,1); put(2,2); get(1); put(3,3); get(2)[null,null,null,1,null,-1]Approaches
1. Array-based naive
Reorder array on access; O(n) per op.
- Time
- O(n) per op
- Space
- O(n)
class LRUCache { constructor(c){this.c=c;this.a=[];} get(k){const i=this.a.findIndex(x=>x[0]===k);if(i<0)return -1;const e=this.a.splice(i,1)[0];this.a.push(e);return e[1];} put(k,v){const i=this.a.findIndex(x=>x[0]===k);if(i>=0)this.a.splice(i,1);this.a.push([k,v]);if(this.a.length>this.c)this.a.shift();} }Tradeoff:
2. Map preserves insertion order
Use JavaScript Map iteration order — delete+set to refresh LRU position.
- Time
- O(1) per op
- Space
- O(capacity)
class LRUCache {
constructor(capacity) { this.cap = capacity; this.map = new Map(); }
get(key) {
if (!this.map.has(key)) return -1;
const v = this.map.get(key);
this.map.delete(key); this.map.set(key, v);
return v;
}
put(key, val) {
if (this.map.has(key)) this.map.delete(key);
this.map.set(key, val);
if (this.map.size > this.cap) this.map.delete(this.map.keys().next().value);
}
}Tradeoff:
Adyen-specific tips
Adyen interviewers grade LRU because their idempotency-key store sits right behind a bounded cache — they expect the O(1) per-op contract and a one-line eviction rule.
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