Paano ka na ze ke sanya cache LRU tare da O(1) get da put?

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**Cache LRU (Least Recently Used)** shine abin da zai cire abin da ba a ɗaukar lura a cikin girma na gida lokacin da ya kai ƙarfin sa. Tsarin gida yana haɗa **hash map** (O(1) bincike) tare da **doubly linked list** (O(1) sake-jera), wanda ke ba **O(1) get da put**. ## The two-structure design ```text HashMap: key -> node DLL (recency order): MRU <-> ... <-> LRU get/put: map finds node move touched node to front (MRU) in O(1); DLL splices it evict the tail (LRU) when full to the front in O(1) ``` ## Code ```python from collections import OrderedDict class LRUCache: def __init__(self, capacity): self.cap = capacity self.cache = OrderedDict() # preserves access order def get(self, key): # O(1) if key not in self.cache: return -1 self.cache.move_to_end(key) # mark as most recently used return self.cache[key] def put(self, key, value): # O(1) if key in self.cache: self.cache.move_to_end(key) self.cache[key] = value if len(self.cache) > self.cap: self.cache.popitem(last=False) # evict least recently used ``` `OrderedDict` shine jaye hash map kuma doubly linked list, daidai jiya tsarin da aka bayyana a sama. | Operation | Time | |---|---| | get | O(1) | | put | O(1) | | space | O(capacity) | ## Why the doubly linked list Doubbly linked list yana ba ka damar in **kawai node daga tsakiya a O(1)** (kana riƙe `prev`/`next` nasu), wanda ba zai yuwu ba tare da singly linked list ko array ba tare da jujjuyawar O(n). ## Dalilin da ya sa haka ya ma'anya Caches LRU suna zaune gida gida da abokin waje, sabar web, da CPUs don ɗauke bakin jini mishi da sauri da alaƙa ƙarfin ƙarfin. Aikin nan shine babbar samari na jira saboda yana bada ƙoƙarin haɗe tsarin biyu sannan *kowane gida na jumloli* — muhimmiyar ƙirar tsarin fasaha.