LRU Cache LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Link for the ProblemLRU Cache– LeetCode Problem

LRU Cache– LeetCode Problem

Problem:

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 <= 104
  • 0 <= value <= 105
  • At most 2 * 105 calls will be made to get and put.
LRU Cache– LeetCode Solutions
LRU Cache Solution in C++:
struct Node {
  int key;
  int value;
};

class LRUCache {
 public:
  LRUCache(int capacity) : capacity(capacity) {}

  int get(int key) {
    if (!keyToIterator.count(key))
      return -1;

    const auto& it = keyToIterator[key];
    cache.splice(begin(cache), cache, it);  // move it to front
    return it->value;
  }

  void put(int key, int value) {
    if (keyToIterator.count(key)) {  // no capacity issue, just update the value
      const auto& it = keyToIterator[key];
      cache.splice(begin(cache), cache, it);  // move it to front
      it->value = value;
      return;
    }

    if (cache.size() == capacity) {  // check the capacity
      const Node& lastNode = cache.back();
      keyToIterator.erase(lastNode.key);  // that's why we store key in `Node`
      cache.pop_back();
    }

    cache.push_front({key, value});
    keyToIterator[key] = begin(cache);
  }

 private:
  const int capacity;
  list<Node> cache;
  unordered_map<int, list<Node>::iterator> keyToIterator;
};
LRU Cache Solution in Java:
class Node {
  public int key;
  public int value;

  public Node(int key, int value) {
    this.key = key;
    this.value = value;
  }
}

class LRUCache {
  public LRUCache(int capacity) {
    this.capacity = capacity;
  }

  public int get(int key) {
    if (!keyToNode.containsKey(key))
      return -1;

    Node node = keyToNode.get(key);
    cache.remove(node);
    cache.add(node);
    return node.value;
  }

  public void put(int key, int value) {
    if (keyToNode.containsKey(key)) {
      keyToNode.get(key).value = value;
      get(key);
      return;
    }

    if (cache.size() == capacity) {
      Node lastNode = cache.iterator().next();
      cache.remove(lastNode);
      keyToNode.remove(lastNode.key);
    }

    Node node = new Node(key, value);
    cache.add(node);
    keyToNode.put(key, node);
  }

  private int capacity;
  private Set<Node> cache = new LinkedHashSet<>();
  private Map<Integer, Node> keyToNode = new HashMap<>();
}
LRU Cache Solution in Python:
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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