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feat: refactored docs structure
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---
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title: '146. LRU Cache'
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description: Design a data structure that follows the constraints of a Least Recently Used (LRU) cache
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icon: dot
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topics:
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- { name: "Hash Table", slug: "hash-table" }
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- { name: "Linked List", slug: "linked-list" }
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- { name: "Design", slug: "design" }
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- { name: "Doubly-Linked List", slug: "doubly-linked-list" }
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video: "7ABFKPK2hD4"
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issue: "https://github.com/prdlk/leetcode/issues/87"
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sidebar:
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label: 'LRU Cache'
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badge: 'Medium'
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---
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### Example 1:
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- Input: ``
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- Output: ``
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- Explanation: 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`
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### Constraints:
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- `1 <= capacity <= 3000`
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- `0 <= key <= 10^4`
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- `0 <= value <= 10^5`
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- At most 2 * 10^5 calls will be made to get and put.
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## Approach
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```mermaid
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flowchart TD
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subgraph state["state — dict for O(1) lookup, doubly linked list for O(1) reorder"]
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H["head — least recent"] --- N["... nodes ..."] --- T["tail — most recent"]
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end
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subgraph get["get(key)"]
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G1{"key in cache?"} -- no --> G2(["return -1"])
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G1 -- yes --> G3["_remove_node then _add_node — move it beside the tail"]
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G3 --> G4(["return node.value"])
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end
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subgraph put["put(key, value)"]
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P1{"key in cache?"} -- yes --> P2["update value, _remove_node then _add_node"]
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P1 -- no --> P3["new Node, cache[key] = node, _add_node"]
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P3 --> P4{"len(cache) > capacity?"}
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P4 -- yes --> P5["evict head.next — the least recent — and del cache[lru.key]"]
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P4 -- no --> P6(["done"])
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P2 --> P6
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P5 --> P6
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end
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state --> get
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state --> put
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```
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## Solution
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```py
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class Node:
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def __init__(self, key=0, value=0):
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self.key = key
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self.value = value
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self.prev = None
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self.next = None
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class LRUCache:
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def __init__(self, capacity: int):
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self.capacity = capacity
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self.cache = {}
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self.head = Node()
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self.tail = Node()
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self.head.next = self.tail # pyright: ignore[reportAttributeAccessIssue]
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self.tail.prev = self.head # pyright: ignore[reportAttributeAccessIssue]
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def _remove_node(self, node):
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prev_node = node.prev
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next_node = node.next
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prev_node.next = next_node
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next_node.prev = prev_node
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def _add_node(self, node):
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# Insert just before the tail (most recent at end)
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node.next = self.tail
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node.prev = self.tail.prev
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self.tail.prev.next = node # pyright: ignore[reportAttributeAccessIssue]
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self.tail.prev = node
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def get(self, key: int) -> int:
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if key not in self.cache:
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return -1
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node = self.cache[key]
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self._remove_node(node)
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self._add_node(node)
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return node.value
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def put(self, key: int, value: int) -> None:
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if key in self.cache:
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node = self.cache[key]
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node.value = value
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self._remove_node(node)
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self._add_node(node)
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else:
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new_node = Node(key, value)
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self.cache[key] = new_node
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self._add_node(new_node)
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if len(self.cache) > self.capacity:
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lru = self.head.next # Oldest node is next to the head
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self._remove_node(lru)
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del self.cache[lru.key] # pyright: ignore[reportAttributeAccessIssue]
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# Your LRUCache object will be instantiated and called as such:
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# obj = LRUCache(capacity)
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# param_1 = obj.get(key)
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# obj.put(key,value)
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```
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@@ -1,76 +0,0 @@
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---
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title: '2. Add Two Numbers'
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description: You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list
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icon: dot
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topics:
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- { name: "Linked List", slug: "linked-list" }
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- { name: "Math", slug: "math" }
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- { name: "Recursion", slug: "recursion" }
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video: "wgFPrzTjm7s"
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issue: "https://github.com/prdlk/leetcode/issues/88"
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sidebar:
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label: 'Add Two Numbers'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `l1 = [2,4,3], l2 = [5,6,4]`
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- Output: `[7,0,8]`
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- Explanation: `342 + 465 = 807`.
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### Example 2:
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- Input: `l1 = [0], l2 = [0]`
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- Output: `[0]`
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### Example 3:
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- Input: `l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]`
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- Output: `[8,9,9,9,0,0,0,1]`
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### Constraints:
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- The number of nodes in each linked list is in the range [1, 100].
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- `0 <= Node.val <= 9`
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- It is guaranteed that the list represents a number that does not have leading zeros.
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## Approach
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```mermaid
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flowchart TD
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S(["addTwoNumbers(l1, l2)"]) --> I["dummy = ListNode(), carry = 0, curr = dummy — dummy.next will be the head"]
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I --> W{"l1 or l2 or carry?"}
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W -- no --> E(["return dummy.next"])
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W -- yes --> A["s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry — a missing digit counts as 0"]
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A --> D["carry, val = divmod(s, 10)"]
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D --> N["curr.next = ListNode(val), curr = curr.next"]
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N --> V["l1 = l1.next if l1 else None, l2 = l2.next if l2 else None"]
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V --> W
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```
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## Solution
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```py
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def addTwoNumbers(
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self, l1: Optional[ListNode], l2: Optional[ListNode]
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) -> Optional[ListNode]:
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dummy = ListNode()
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carry = 0
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curr = dummy
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while l1 or l2 or carry:
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s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry
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carry, val = divmod(s, 10)
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curr.next = ListNode(val)
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curr = curr.next
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l1 = l1.next if l1 else None
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l2 = l2.next if l2 else None
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return dummy.next
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```
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@@ -1,5 +0,0 @@
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---
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title: Hybrid Structures
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description: 'Two structures are combined, such as a hash map and a linked list, to get the best properties of each.'
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hidden: true
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---
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@@ -1,3 +0,0 @@
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import { defineMeta } from "blume";
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export default defineMeta({ title: "Hybrid Structures", order: 9, icon: "combine", display: "page" });
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