# LRU Cache

• `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.
`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]ExplanationLRUCache 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 1lRUCache.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 3lRUCache.get(4);    // return 4`
• `1 <= capacity <= 3000`
• `0 <= key <= 3000`
• `0 <= value <= 104`
• At most `3 * 104` calls will be made to `get` and `put`.
• Time complexity : O(1) both for `put` and `get`.
• Space complexity : O(capacity) since the space is used only for a hashmap and double linked list with at most `capacity + 1` elements.

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Full Stack Programmer, love to solve problem’s during free time.

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## Jaydeep

Full Stack Programmer, love to solve problem’s during free time.