What is the brute force approach and complexity for LRU Cache?

Vector of (key, value) from most to least recent; linear search on every call with time complexity O(C) and space complexity O(C).

What is the optimal approach and complexity for LRU Cache?

List keeps recency order; map key → list node; splice to front, evict from back with optimal time complexity O(1) and space complexity O(C).

LRU Cache

Q: How do you identify the pattern for LRU Cache?

Look for key signals in the prompt: "get and put in O(1)" indicates No scanning → a hash map to find keys.; "least recently used" indicates We need an order of use, with the oldest easy to remove → a list with an old end.; "a use moves the key to the front" indicates Items must jump from the middle to the front → a doubly linked list, not a queue or vector..

Medium·LeetCode #146 Pro OnlyMedium

LeetCode

Problem Statement

Build a class LRUCache: a key–value store with a fixed size limit. When it is full and a new key arrives, it throws out the key that was used longest ago (the least recently used, "LRU").

LRUCache(int capacity) creates an empty cache that can hold at most capacity keys.
int get(int key) returns the value of key, or -1 if it is not stored. A successful get counts as a use of that key.
void put(int key, int value) stores value under key. If the key already exists, its value is replaced. Otherwise it is added, and if that makes the cache hold more than capacity keys, the least recently used key is removed first. A put also counts as a use.

Both get and put must run in O(1) average time.

Input

The capacity, then a sequence of get(key) and put(key, value) calls.

Output

For each get, the stored value or -1. put returns nothing.

Example 1

Input: capacity = 2 put(1, 1), put(2, 2), get(1), put(3, 3), get(2), put(4, 4), get(1), get(3), get(4)

Output: 1, -1, -1, 3, 4 (the results of the five get calls)

After get(1), key 1 is the most recent, so put(3, 3) evicts key 2. Then key 1 is the oldest, so put(4, 4) evicts key 1. Keys 3 and 4 remain.

Constraints

• 1 ≤ capacity ≤ 3000
• 0 ≤ key ≤ 10^4
• 0 ≤ value ≤ 10^5
• At most 2 · 10^5 calls to get and put

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“get and put in O(1)”

→ No scanning → a hash map to find keys.

“least recently used”

→ We need an order of use, with the oldest easy to remove → a list with an old end.

“a use moves the key to the front”

→ Items must jump from the middle to the front → a doubly linked list, not a queue or vector.

Approach Complexity Ladder

Brute Force — vector in recency orderO(C)O(C)

Vector of (key, value) from most to least recent; linear search on every call

Optimal — hash map + doubly linked listO(1)O(C)

List keeps recency order; map key → list node; splice to front, evict from back

Brute Force Intuition

Keep a vector<pair<int,int>> ordered from most recently used (index 0) to least recently used (the back). get and put search the vector for the key. On a hit, erase the pair and re-insert it at the front. To evict, pop_back.

Recognition rule for this sub-pattern:

You must support operations in O(1); combine a hash map with another structure (list, array) so each one covers the other's weakness.

Remember this · Arrays & Hashing

Trigger

"Have I seen X before?", pairs that add up to a target, grouping things that look alike, counting occurrences.

Think

Instead of scanning again for each element, can I remember earlier elements in a map?

Key question

What exactly do I store as the key, and what as the value?

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Beyond The LeetCode Problem

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LLD Machine Coding

In-Memory LRU / LFU Cache

Hash map combined with doubly linked list for O(1) eviction and lookup.

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STL unordered_map Internals & Rehashing

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Insert Delete GetRandom O(1)LFU Cache