feat(work): add LRU Cache solution

work/1/Medium/Hash Table/146.lru-cache.py

@@ -0,0 +1,124 @@
+"""
+146. LRU Cache
+Difficulty: Medium
+https://leetcode.com/problems/lru-cache/
+
+──────────────────────────────────────────────────
+
+Design a data structure that follows the constraints of a Least
+Recently Used (LRU) cache.
+
+Implement the LRUCache class:
+
+• 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.
+
+The functions get and put must each run in O(1) average time
+complexity.
+
+
+
+Example 1:
+
+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]
+
+Explanation
+LRUCache 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 1
+lRUCache.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 3
+lRUCache.get(4);    // return 4
+
+
+
+Constraints:
+
+        • 1 <= capacity <= 3000
+
+        • 0 <= key <= 10^4
+
+        • 0 <= value <= 10^5
+
+        • At most 2 * 10^5 calls will be made to get and put.
+"""
+
+
+class Node:
+    def __init__(self, key=0, value=0):
+        self.key = key
+        self.value = value
+        self.prev = None
+        self.next = None
+
+
+class LRUCache:
+    def __init__(self, capacity: int):
+        self.capacity = capacity
+        self.cache = {}
+        self.head = Node()
+        self.tail = Node()
+        self.head.next = self.tail  # pyright: ignore[reportAttributeAccessIssue]
+        self.tail.prev = self.head  # pyright: ignore[reportAttributeAccessIssue]
+
+    def _remove_node(self, node):
+        prev_node = node.prev
+        next_node = node.next
+        prev_node.next = next_node
+        next_node.prev = prev_node
+
+    def _add_node(self, node):
+        node.prev = self.head
+        node.next = self.head.next
+        self.head.next.prev = node  # pyright: ignore[reportAttributeAccessIssue]
+        self.head.next = node
+
+    def get(self, key: int) -> int:
+        if key not in self.cache:
+            return -1
+
+        node = self.cache[key]
+        self._remove_node(node)
+        self._add_node(node)
+        return node.value
+
+    def put(self, key: int, value: int) -> None:
+        if key in self.cache:
+            node = self.cache[key]
+            node.value = value
+            self._remove_node(node)
+            self._add_node(node)
+        else:
+            new_node = Node(key, value)
+            self.cache[key] = new_node
+            self._add_node(new_node)
+
+            if len(self.cache) > self.capacity:
+                lru = self.tail.prev
+                self._remove_node(lru)
+                del self.cache[lru.key]  # pyright: ignore[reportAttributeAccessIssue]
+
+
+# Your LRUCache object will be instantiated and called as such:
+# obj = LRUCache(capacity)
+# param_1 = obj.get(key)
+# obj.put(key,value)