# 146. LRU Cache LeetCode Solution | Easy Approach

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LRU Cache 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.

Example 1:

```Input
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[, [1, 1], [2, 2], , [3, 3], , [4, 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 <= 104`
• `0 <= value <= 105`
• At most 2` * 105` calls will be made to `get` and `put`.

Time: O(n)
Space: O(n)

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