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Problem solution · Python

LFU Cache

LFU 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
65 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 LFU 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

  • 65 lines of Python from the credited upstream file 460.py.
  • The implementation visibly relies on sequence storage.
  • 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 codeLFU Cache · PythonPython
Use this to learn the idea, then write your own version.
class Node:  def __init__(self, key: int, value: int, freq: int, it):    self.key = key    self.value = value    self.freq = freq    self.it = it  class LFUCache:  def __init__(self, capacity: int):    self.capacity = capacity    self.minFreq = 0    self.keyToNode = {}    self.freqToList = {}   def get(self, key: int) -> int:    if key not in self.keyToNode:      return -1     node = self.keyToNode[key]    self._touch(node)    return node.value   def put(self, key: int, value: int) -> None:    if self.capacity == 0:      return     if key in self.keyToNode:      node = self.keyToNode[key]      node.value = value      self._touch(node)      return     if len(self.keyToNode) == self.capacity:      # Evict an LRU key from `minFreq` list      keyToEvict = self.freqToList[self.minFreq][-1]      self.freqToList[self.minFreq].pop()      del self.keyToNode[keyToEvict]     self.minFreq = 1    if 1 not in self.freqToList:      self.freqToList[1] = []    self.freqToList[1].insert(0, key)    self.keyToNode[key] = Node(key, value, 1, 0)  # Using index 0 as iterator   def _touch(self, node: Node) -> None:    # Update the node's frequency    prevFreq = node.freq    node.freq += 1    newFreq = node.freq     # Remove the key from `prevFreq`'s list    self.freqToList[prevFreq].remove(node.key)    if not self.freqToList[prevFreq]:      del self.freqToList[prevFreq]      # Update `minFreq` if needed      if prevFreq == self.minFreq:        self.minFreq += 1     # Insert the key to the front of `newFreq`'s list    if newFreq not in self.freqToList:      self.freqToList[newFreq] = []    self.freqToList[newFreq].insert(0, node.key)    node.it = 0  # Using index 0 as iterator

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