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Problem solution · Python

LRU Cache

LRU Cache: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
52 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For LRU Cache, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 52 lines of Python from the credited upstream file 146.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeLRU Cache · PythonPython
Use this to learn the idea, then write your own version.
class Node:  def __init__(self, key: int, value: int):    self.key = key    self.value = value    self.prev = None    self.next = None  class LRUCache:  def __init__(self, capacity: int):    self.capacity = capacity    self.keyToNode = {}    self.head = Node(-1, -1)    self.tail = Node(-1, -1)    self.join(self.head, self.tail)   def get(self, key: int) -> int:    if key not in self.keyToNode:      return -1     node = self.keyToNode[key]    self.remove(node)    self.moveToHead(node)    return node.value   def put(self, key: int, value: int) -> None:    if key in self.keyToNode:      node = self.keyToNode[key]      node.value = value      self.remove(node)      self.moveToHead(node)      return     if len(self.keyToNode) == self.capacity:      lastNode = self.tail.prev      del self.keyToNode[lastNode.key]      self.remove(lastNode)     self.moveToHead(Node(key, value))    self.keyToNode[key] = self.head.next   def join(self, node1: Node, node2: Node):    node1.next = node2    node2.prev = node1   def moveToHead(self, node: Node):    self.join(node, self.head.next)    self.join(self.head, node)   def remove(self, node: Node):    self.join(node.prev, node.next)

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