Web Reference: When the cache reaches its capacity, it should invalidate and remove the least frequently used key before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), the least recently used key would be invalidated. In-depth solution and explanation for LeetCode 460. LFU Cache in Python, Java, C++ and more. Intuitions, example walk through, and complexity analysis. Better than official and forum solutions. To solve LeetCode 460: LFU Cache in Python, we need a data structure supporting O (1) get and put, tracking frequency and recency, and evicting the least frequently used (LFU) item with LRU tiebreaking.
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