https://dmytros.blog/tags/linked_list/ Recent content in Linked_List on Dmytro's Blog https://dmytros.blog/images/papermod-cover.png https://dmytros.blog/images/papermod-cover.png Hugo en Fri, 18 Sep 2026 23:10:02 +0300 https://dmytros.blog/posts/leetcode-lfu-cache/ Fri, 18 Sep 2026 23:10:02 +0300 https://dmytros.blog/posts/leetcode-lfu-cache/ <h3 id="the-problem">The problem</h3> <p>Design and implement a data structure for a <a href="https://en.wikipedia.org/wiki/Least_frequently_used">Least Frequently Used (LFU)</a> cache.</p> <p>Implement the <code>LFUCache</code> class:</p> <ul> <li><code>LFUCache(int capacity)</code> Initializes the object with the <code>capacity</code> of the data structure.</li> <li><code>int get(int key)</code> Gets the value of the <code>key</code> if the <code>key</code> exists in the cache. Otherwise, returns <code>-1</code>.</li> <li><code>void put(int key, int value)</code> Updates the value of the <code>key</code> if present, or inserts the <code>key</code> if not already present. When the cache reaches its <code>capacity</code>, it should invalidate and remove the <strong>least frequently used</strong> key before inserting a new item. For this problem, when there is a <strong>tie</strong> (i.e., two or more keys with the same frequency), the <strong>least recently used</strong> <code>key</code> would be invalidated.</li> </ul> <p>To determine the least frequently used key, a <strong>use counter</strong> is maintained for each key in the cache. The key with the smallest <strong>use counter</strong> is the least frequently used key.</p>