LRU (Least Recently Used) Cache discards the least recently used items first. This algorithm requires keeping track of what was used when, which is expensive if one wants to make sure the algorithm always discards the least recently used item. General implementations of this technique require keeping “age bits” for cache-lines and track the “Least Recently Used” cache-line based on age-bits.
Our problem statement is to design and implement a data structure for Least Recently Used (LRU) cache.
It should support the following operations: get and put.
* get(key) – Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
* put(key, value) – Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
The cache is always initialized with positive capacity.
Input/Output : LRUCache cache = new LRUCache( 2 /* capacity */ ); cache.put(1, 1); cache.put(2, 2); cache.get(1); // returns 1 cache.put(3, 3); // evicts key 2 cache.get(2); // returns -1 (not found) cache.put(4, 4); // evicts key 1 cache.get(1); // returns -1 (not found) cache.get(3); // returns 3 cache.get(4); // returns 4
Our solution is to use the power of OrderedDict from collections module which keep order of insertion of keys and we can change that order if required. The best part is all operations have O(1) time complexity.
We maintain our OrderedDict in such a way that the order shows how recently they were used. In the beginning, we will have least recently used and in the end, most recently used.
If any update OR query is made to a key it moves to the end (most recently used). If anything is added, it is added at the end (most recently used/added)
For get(key): we return the value of the key that is queried in O(1) and return -1 if we don’t find the key in out dict/cache. And also move the key to the end to show that it was recently used.
For put(key, value): first, we add/ update the key by conventional methods. And also move the key to the end to show that it was recently used. But here we will also check whether the length of our ordered dictionary has exceeded our capacity, If so we remove the first key (least recently used)
OrderedDict([(1, 1)]) OrderedDict([(1, 1), (2, 2)]) OrderedDict([(2, 2), (1, 1)]) OrderedDict([(1, 1), (3, 3)]) OrderedDict([(1, 1), (3, 3)]) OrderedDict([(3, 3), (4, 4)]) OrderedDict([(3, 3), (4, 4)]) OrderedDict([(4, 4), (3, 3)]) OrderedDict([(3, 3), (4, 4)])
Time Complexity :O(1)
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