Design a data structure for LRU Cache
Design a data structure for LRU Cache. It should support the following operations: get and set.
- get(key) – Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
- set(key, value) – Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
Examples:
// Let's say we have a LRU cache of capacity 2.
LRUCache cache = new LRUCache(2);
cache.set(1, 10); // it will store a key (1) with value 10 in the cache.
cache.set(2, 20); // it will store a key (2) with value 20 in the cache.
cache.get(1); // returns 10
cache.set(3, 30); // evicts key 2 and store a key (3) with value 30 in the cache.
cache.get(2); // returns -1 (not found)
cache.set(4, 40); // evicts key 1 and store a key (4) with value 40 in the cache.
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 30
cache.get(4); // returns 40
Asked In: Adobe, Hike and many more companies.
Solution:
Brute-force Approach: We will keep an array of Nodes and each node will contain the following information:
Implementation:
C++
struct Node { int key; int value; // It shows the time at which the key is stored. // We will use the timeStamp to find out the // least recently used (LRU) node. int timeStamp; Node( int _key, int _value) { key = _key; value = _value; // currentTimeStamp from system timeStamp = currentTimeStamp; } }; // This code is contributed by subham348 |
Java
class Node { int key; int value; // it shows the time at which the key is stored. // We will use the timeStamp to find out the // least recently used (LRU) node. int timeStamp; public Node( int key, int value) { this .key = key; this .value = value; // currentTimeStamp from system this .timeStamp = currentTimeStamp; } } |
Python3
# using time module import time class Node: # timestamp shows the time at which the key is stored. # We will use the timeStamp to find out the # least recently used (LRU) node. def __init__( self , key, value): self .key = key self .value = value # currentTimeStamp from system self .timeStamp = time.time() # This code is contributed by jainlovely450. |
C#
using System; public class Node { public int Key { get ; set ; } public int Value { get ; set ; } // It shows the time at which the key is stored. // We will use the timeStamp to find out the // least recently used (LRU) node. public DateTime TimeStamp { get ; set ; } public Node( int _key, int _value) { Key = _key; Value = _value; // currentTimeStamp from system TimeStamp = DateTime.Now; } } class Program { static void Main() { // Example usage: Node myNode = new Node(1, 42); Console.WriteLine($ "Key: {myNode.Key}, Value: {myNode.Value}, TimeStamp: {myNode.TimeStamp}" ); } } // This code is contributed by shivamgupta310570 |
Javascript
class Node { constructor(key, value) { this .key = key; this .value = value; // It shows the time at which the key is stored. // We will use the timeStamp to find out the // least recently used (LRU) node. this .timeStamp = Date.now(); // Current timestamp in milliseconds } } // Example usage: const node = new Node(1, "Value1" ); console.log(node.key); // Output: 1 console.log(node.value); // Output: "Value1" console.log(node.timeStamp); // Output: Current timestamp |
The size of the array will be equal to the given capacity of cache.
- For get(int key): We can simply iterate over the array and compare the key of each node with the given key and return the value stored in the node for that key. If we don’t find any such node, return simply -1.
Time Complexity: O(n) - For set(int key, int value): If the array if full, we have to delete one node from the array. To find the LRU node, we will iterate through the array and find the node with least timeStamp value. We will simply insert the new node (with new key and value) at the place of the LRU node.
If the array is not full, we can simply insert a new node in the array at the last current index of the array.
Time Complexity: O(n)
Optimized Approach: The key to solve this problem is using a double linked list which enables us to quickly move nodes.
The LRU cache is a hash map of keys and double linked nodes. The hash map makes the time of get() to be O(1). The list of double linked nodes make the nodes adding/removal operations O(1).
Code using Doubly Linked List and HashMap:
C++
#include <bits/stdc++.h> using namespace std; class LRUCache{ public : class node { public : int key; int value; node * prev; node * next; node( int _key, int _value) { key = _key; value = _value; } }; node* head = new node(-1, -1); node* tail = new node(-1, -1); int cap; map< int , node *> m; // Constructor for initializing the // cache capacity with the given value. LRUCache( int capacity) { cap = capacity; head->next = tail; tail->prev = head; } void addnode(node * temp) { node * dummy = head->next; head->next = temp; temp->prev = head; temp->next = dummy; dummy->prev = temp; } void deletenode(node * temp) { node * delnext = temp->next; node * delprev = temp->prev; delnext->prev = delprev; delprev->next = delnext; } // This method works in O(1) int get( int key) { if (m.find(key) != m.end()) { node * res = m[key]; m.erase(key); int ans = res->value; deletenode(res); addnode(res); m[key] = head->next; cout << "Got the value : " << ans << " for the key: " << key << "\n" ; return ans; } cout << "Did not get any value for the key: " << key << "\n" ; return -1; } // This method works in O(1) void set( int key, int value) { cout << "Going to set the (key, value) : (" << key << ", " << value << ")" << "\n" ; if (m.find(key) != m.end()) { node * exist = m[key]; m.erase(key); deletenode(exist); } if (m.size() == cap) { m.erase(tail->prev->key); deletenode(tail->prev); } addnode( new node(key, value)); m[key] = head->next; } }; // Driver code int main() { cout << "Going to test the LRU " << "Cache Implementation\n" ; LRUCache * cache = new LRUCache(2); // It will store a key (1) with value // 10 in the cache. cache->set(1, 10); // It will store a key (1) with value 10 in the // cache. cache->set(2, 20); cout << "Value for the key: 1 is " << cache->get(1) << "\n" ; // returns 10 // Evicts key 2 and store a key (3) with // value 30 in the cache. cache->set(3, 30); cout << "Value for the key: 2 is " << cache->get(2) << "\n" ; // returns -1 (not found) // Evicts key 1 and store a key (4) with // value 40 in the cache. cache->set(4, 40); cout << "Value for the key: 1 is " << cache->get(1) << "\n" ; // returns -1 (not found) cout << "Value for the key: 3 is " << cache->get(3) << "\n" ; // returns 30 cout << "Value for the key: 4 is " << cache->get(4) << "\n" ; // return 40 return 0; } // This code is contributed by CoderSaty |
Java
import java.util.HashMap; class Node { int key; int value; Node pre; Node next; public Node( int key, int value) { this .key = key; this .value = value; } } class LRUCache { private HashMap<Integer, Node> map; private int capacity, count; private Node head, tail; public LRUCache( int capacity) { this .capacity = capacity; map = new HashMap<>(); head = new Node( 0 , 0 ); tail = new Node( 0 , 0 ); head.next = tail; tail.pre = head; head.pre = null ; tail.next = null ; count = 0 ; } public void deleteNode(Node node) { node.pre.next = node.next; node.next.pre = node.pre; } public void addToHead(Node node) { node.next = head.next; node.next.pre = node; node.pre = head; head.next = node; } // This method works in O(1) public int get( int key) { if (map.get(key) != null ) { Node node = map.get(key); int result = node.value; deleteNode(node); addToHead(node); System.out.println( "Got the value : " + result + " for the key: " + key); return result; } System.out.println( "Did not get any value" + " for the key: " + key); return - 1 ; } // This method works in O(1) public void set( int key, int value) { System.out.println( "Going to set the (key, " + "value) : (" + key + ", " + value + ")" ); if (map.get(key) != null ) { Node node = map.get(key); node.value = value; deleteNode(node); addToHead(node); } else { Node node = new Node(key, value); map.put(key, node); if (count < capacity) { count++; addToHead(node); } else { map.remove(tail.pre.key); deleteNode(tail.pre); addToHead(node); } } } } public class TestLRUCache { public static void main(String[] args) { System.out.println( "Going to test the LRU " + " Cache Implementation" ); LRUCache cache = new LRUCache( 2 ); // it will store a key (1) with value // 10 in the cache. cache.set( 1 , 10 ); // it will store a key (1) with value 10 in the // cache. cache.set( 2 , 20 ); System.out.println( "Value for the key: 1 is " + cache.get( 1 )); // returns 10 // evicts key 2 and store a key (3) with // value 30 in the cache. cache.set( 3 , 30 ); System.out.println( "Value for the key: 2 is " + cache.get( 2 )); // returns -1 (not found) // evicts key 1 and store a key (4) with // value 40 in the cache. cache.set( 4 , 40 ); System.out.println( "Value for the key: 1 is " + cache.get( 1 )); // returns -1 (not found) System.out.println( "Value for the key: 3 is " + cache.get( 3 )); // returns 30 System.out.println( "Value for the key: 4 is " + cache.get( 4 )); // return 40 } } |
Python3
# Class for a Doubly LinkedList Node class DLLNode: def __init__( self , key, val): self .val = val self .key = key self .prev = None self . next = None # LRU cache class class LRUCache: def __init__( self , capacity): # capacity: capacity of cache # Initialize all variable self .capacity = capacity self . map = {} self .head = DLLNode( 0 , 0 ) self .tail = DLLNode( 0 , 0 ) self .head. next = self .tail self .tail.prev = self .head self .count = 0 def deleteNode( self , node): node.prev. next = node. next node. next .prev = node.prev def addToHead( self , node): node. next = self .head. next node. next .prev = node node.prev = self .head self .head. next = node # This method works in O(1) def get( self , key): if key in self . map : node = self . map [key] result = node.val self .deleteNode(node) self .addToHead(node) print ( 'Got the value : {} for the key: {}' . format (result, key)) return result print ( 'Did not get any value for the key: {}' . format (key)) return - 1 # This method works in O(1) def set ( self , key, value): print ( 'going to set the (key, value) : ( {}, {})' . format (key, value)) if key in self . map : node = self . map [key] node.val = value self .deleteNode(node) self .addToHead(node) else : node = DLLNode(key, value) self . map [key] = node if self .count < self .capacity: self .count + = 1 self .addToHead(node) else : del self . map [ self .tail.prev.key] self .deleteNode( self .tail.prev) self .addToHead(node) if __name__ = = '__main__' : print ( 'Going to test the LRU Cache Implementation' ) cache = LRUCache( 2 ) # it will store a key (1) with value # 10 in the cache. cache. set ( 1 , 10 ) # it will store a key (1) with value 10 in the cache. cache. set ( 2 , 20 ) print ( 'Value for the key: 1 is {}' . format (cache.get( 1 ))) # returns 10 # evicts key 2 and store a key (3) with # value 30 in the cache. cache. set ( 3 , 30 ) print ( 'Value for the key: 2 is {}' . format ( cache.get( 2 ))) # returns -1 (not found) # evicts key 1 and store a key (4) with # value 40 in the cache. cache. set ( 4 , 40 ) print ( 'Value for the key: 1 is {}' . format ( cache.get( 1 ))) # returns -1 (not found) print ( 'Value for the key: 3 is {}' . format (cache.get( 3 ))) # returns 30 print ( 'Value for the key: 4 is {}' . format (cache.get( 4 ))) # returns 40 |
C#
using System; using System.Collections.Generic; class Node { public int Key { get ; set ; } public int Value { get ; set ; } public Node Pre { get ; set ; } public Node Next { get ; set ; } public Node( int key, int value) { Key = key; Value = value; } } class LRUCache { private Dictionary< int , Node> map; private int capacity, count; private Node head, tail; public LRUCache( int capacity) { this .capacity = capacity; map = new Dictionary< int , Node>(); head = new Node(0, 0); tail = new Node(0, 0); head.Next = tail; tail.Pre = head; head.Pre = null ; tail.Next = null ; count = 0; } public void DeleteNode(Node node) { node.Pre.Next = node.Next; node.Next.Pre = node.Pre; } public void AddToHead(Node node) { node.Next = head.Next; node.Next.Pre = node; node.Pre = head; head.Next = node; } // This method works in O(1) public int Get( int key) { if (map.ContainsKey(key)) { Node node = map[key]; int result = node.Value; DeleteNode(node); AddToHead(node); Console.WriteLine( "Got the value: " + result + " for the key: " + key); return result; } Console.WriteLine( "Did not get any value for the key: " + key); return -1; } // This method works in O(1) public void Set( int key, int value) { Console.WriteLine( "Going to set the (key, value): (" + key + ", " + value + ")" ); if (map.ContainsKey(key)) { Node node = map[key]; node.Value = value; DeleteNode(node); AddToHead(node); } else { Node node = new Node(key, value); map[key] = node; if (count < capacity) { count++; AddToHead(node); } else { map.Remove(tail.Pre.Key); DeleteNode(tail.Pre); AddToHead(node); } } } } public class TestLRUCache { public static void Main( string [] args) { Console.WriteLine( "Going to test the LRU Cache Implementation" ); LRUCache cache = new LRUCache(2); // It will store a key (1) with value // 10 in the cache. cache.Set(1, 10); // It will store a key (1) with value 10 in the // cache. cache.Set(2, 20); Console.WriteLine( "Value for the key: 1 is " + cache.Get(1)); // Returns 10 // Evicts key 2 and store a key (3) with // value 30 in the cache. cache.Set(3, 30); Console.WriteLine( "Value for the key: 2 is " + cache.Get(2)); // Returns -1 (not found) // Evicts key 1 and store a key (4) with // value 40 in the cache. cache.Set(4, 40); Console.WriteLine( "Value for the key: 1 is " + cache.Get(1)); // Returns -1 (not found) Console.WriteLine( "Value for the key: 3 is " + cache.Get(3)); // Returns 30 Console.WriteLine( "Value for the key: 4 is " + cache.Get(4)); // Return 40 } } |
Javascript
class LRUCache { constructor(capacity) { // Initialize the LRUCache with given capacity this .cap = capacity; // Create dummy head and tail nodes for the doubly linked list this .head = new Node(-1, -1); this .tail = new Node(-1, -1); this .head.next = this .tail; this .tail.prev = this .head; // Map to store key-node pairs for constant time access this .m = new Map(); } // Add a new node to the beginning of the doubly linked list addNode(temp) { const dummy = this .head.next; this .head.next = temp; temp.prev = this .head; temp.next = dummy; dummy.prev = temp; } // Delete a node from the doubly linked list deleteNode(temp) { const delNext = temp.next; const delPrev = temp.prev; delNext.prev = delPrev; delPrev.next = delNext; } // Get the value for a given key get(key) { if ( this .m.has(key)) { const res = this .m.get(key); this .m. delete (key); const ans = res.value; this .deleteNode(res); this .addNode(res); this .m.set(key, this .head.next); // Log the result console.log(`Got the value: ${ans} for the key: ${key}`); return ans; } // Log if the key is not found console.log(`Did not get any value for the key: ${key}`); return -1; } // Set the value for a given key set(key, value) { // Log the key-value pair being set console.log(`Going to set the (key, value): (${key}, ${value})`); if ( this .m.has(key)) { // If key exists, delete the node and remove it from the map const exist = this .m.get(key); this .m. delete (key); this .deleteNode(exist); } if ( this .m.size === this .cap) { // If capacity is reached, evict the least recently used node this .m. delete ( this .tail.prev.key); this .deleteNode( this .tail.prev); } // Add a new node to the doubly linked list and update the map this .addNode( new Node(key, value)); this .m.set(key, this .head.next); } } // Node class representing a node in the doubly linked list class Node { constructor(key, value) { this .key = key; this .value = value; this .prev = null ; this .next = null ; } } // Driver code console.log( "Going to test the LRU Cache Implementation" ); const cache = new LRUCache(2); // Test the LRUCache with set and get operations cache.set(1, 10); cache.set(2, 20); console.log(`Value for the key: 1 is ${cache.get(1)}`); // returns 10 cache.set(3, 30); console.log(`Value for the key: 2 is ${cache.get(2)}`); // returns -1 (not found) cache.set(4, 40); console.log(`Value for the key: 1 is ${cache.get(1)}`); // returns -1 (not found) console.log(`Value for the key: 3 is ${cache.get(3)}`); // returns 30 console.log(`Value for the key: 4 is ${cache.get(4)}`); // return 40 |
Going to test the LRU Cache Implementation Going to set the (key, value) : (1, 10) Going to set the (key, value) : (2, 20) Got the value : 10 for the key: 1 Value for the key: 1 is 10 Going to set the (key, value) : (3, 30) Did not get any value for the key: 2 Value for the key: 2 is -1 Going to set the (key, value) : (4, 40) Did not get any value for the key: 1 Value for the key: 1 is -1 Got the value : 30 for the key: 3 Value for the key: 3 is 30 Got the value : 40 for the key: 4 Value for the key: 4 is 40
Another implementation in Java using LinkedHashMap:
removeEldestEntry() is overridden to impose a policy for removing old mappings when size goes beyond capacity.
Implementation:
C++
#include <iostream> #include <unordered_map> class LRUCache { private : std::unordered_map< int , int > map; const int CAPACITY; public : // Constructor to initialize the LRUCache with a given capacity LRUCache( int capacity) : CAPACITY(capacity) {} // This method works in O(1) int get( int key) { std::cout << "Going to get the value for the key: " << key << std::endl; auto it = map.find(key); if (it != map.end()) { // Access the key to update its position in the cache (LRU) map.erase(it); map[key] = it->second; return it->second; } return -1; } // This method works in O(1) void set( int key, int value) { std::cout << "Going to set the (key, value): (" << key << ", " << value << ")" << std::endl; auto it = map.find(key); if (it != map.end()) { map.erase(it); } else if (map.size() >= CAPACITY) { // Remove the least recently used item auto first = map.begin(); map.erase(first); } map[key] = value; } }; int main() { std::cout << "Going to test the LRUCache Implementation" << std::endl; LRUCache cache(2); // It will store a key (1) with value 10 in the cache. cache.set(1, 10); // It will store a key (2) with value 20 in the cache. cache.set(2, 20); std::cout << "Value for the key: 1 is " << cache.get(1) << std::endl; // Returns 10 // Evicts key 2 and stores a key (3) with value 30 in the cache. cache.set(3, 30); std::cout << "Value for the key: 2 is " << cache.get(2) << std::endl; // Returns -1 (not found) // Evicts key 1 and stores a key (4) with value 40 in the cache. cache.set(4, 40); std::cout << "Value for the key: 1 is " << cache.get(1) << std::endl; // Returns -1 (not found) std::cout << "Value for the key: 3 is " << cache.get(3) << std::endl; // Returns 30 std::cout << "Value for the key: 4 is " << cache.get(4) << std::endl; // Returns 40 return 0; } |
Java
import java.util.LinkedHashMap; import java.util.Map; class TestLRUCacheWithLinkedHashMap { public static void main(String[] args) { System.out.println( "Going to test the LRU " + " Cache Implementation" ); LRUCache cache = new LRUCache( 2 ); // it will store a key (1) with value // 10 in the cache. cache.set( 1 , 10 ); // it will store a key (1) with value 10 in the cache. cache.set( 2 , 20 ); System.out.println( "Value for the key: 1 is " + cache.get( 1 )); // returns 10 // evicts key 2 and store a key (3) with // value 30 in the cache. cache.set( 3 , 30 ); System.out.println( "Value for the key: 2 is " + cache.get( 2 )); // returns -1 (not found) // evicts key 1 and store a key (4) with // value 40 in the cache. cache.set( 4 , 40 ); System.out.println( "Value for the key: 1 is " + cache.get( 1 )); // returns -1 (not found) System.out.println( "Value for the key: 3 is " + cache.get( 3 )); // returns 30 System.out.println( "Value for the key: 4 is " + cache.get( 4 )); // return 40 } } class LRUCache { private LinkedHashMap<Integer, Integer> map; private final int CAPACITY; public LRUCache( int capacity) { CAPACITY = capacity; map = new LinkedHashMap<Integer, Integer>(capacity, 0 .75f, true ) { protected boolean removeEldestEntry(Map.Entry eldest) { return size() > CAPACITY; } }; } // This method works in O(1) public int get( int key) { System.out.println( "Going to get the value " + "for the key : " + key); return map.getOrDefault(key, - 1 ); } // This method works in O(1) public void set( int key, int value) { System.out.println( "Going to set the (key, " + "value) : (" + key + ", " + value + ")" ); map.put(key, value); } } |
Python3
from collections import OrderedDict class LRUCache: def __init__( self , capacity): self .CAPACITY = capacity self .cache = OrderedDict() def get( self , key): print (f "Going to get the value for the key: {key}" ) return self .cache.get(key, - 1 ) def set ( self , key, value): print (f "Going to set the (key, value): ({key}, {value})" ) self .cache[key] = value if len ( self .cache) > self .CAPACITY: self .cache.popitem(last = False ) # Remove the least recently used entry if __name__ = = "__main__" : print ( "Going to test the LRU Cache Implementation" ) cache = LRUCache( 2 ) cache. set ( 1 , 10 ) cache. set ( 2 , 20 ) print (f "Value for the key: 1 is {cache.get(1)}" ) # returns 10 cache. set ( 3 , 30 ) print (f "Value for the key: 2 is {cache.get(2)}" ) # returns -1 (not found) cache. set ( 4 , 40 ) print (f "Value for the key: 1 is {cache.get(1)}" ) # returns -1 (not found) print (f "Value for the key: 3 is {cache.get(3)}" ) # returns 30 print (f "Value for the key: 4 is {cache.get(4)}" ) # returns 40 |
C#
using System; using System.Collections.Generic; class TestLRUCacheWithLinkedHashMap { static void Main( string [] args) { Console.WriteLine( "Going to test the LRU " + "Cache Implementation" ); LRUCache cache = new LRUCache(2); // it will store a key (1) with value // 10 in the cache. cache.Set(1, 10); // it will store a key (1) with value 10 in the cache. cache.Set(2, 20); Console.WriteLine( "Value for the key: 1 is " + cache.Get(1)); // returns 10 // evicts key 2 and store a key (3) with // value 30 in the cache. cache.Set(3, 30); Console.WriteLine( "Value for the key: 2 is " + cache.Get(2)); // returns -1 (not found) // evicts key 1 and store a key (4) with // value 40 in the cache. cache.Set(4, 40); Console.WriteLine( "Value for the key: 1 is " + cache.Get(1)); // returns -1 (not found) Console.WriteLine( "Value for the key: 3 is " + cache.Get(3)); // returns 30 Console.WriteLine( "Value for the key: 4 is " + cache.Get(4)); // return 40 } } class LRUCache { private readonly Dictionary< int , int > map; private readonly int CAPACITY; public LRUCache( int capacity) { CAPACITY = capacity; map = new Dictionary< int , int >(capacity); } // This method works in O(1) public int Get( int key) { Console.WriteLine( "Going to get the value " + "for the key : " + key); if (map.TryGetValue(key, out int value)) { // Access the key to update its position in cache (LRU) map.Remove(key); map.Add(key, value); return value; } return -1; } // This method works in O(1) public void Set( int key, int value) { Console.WriteLine( "Going to set the (key, " + "value) : (" + key + ", " + value + ")" ); if (map.ContainsKey(key)) { map.Remove(key); } else if (map.Count >= CAPACITY) { // Remove the least recently used item var first = map.Keys.GetEnumerator(); first.MoveNext(); map.Remove(first.Current); } map.Add(key, value); } } |
Javascript
class LRUCache { constructor(capacity) { this .CAPACITY = capacity; this .map = new Map(); } // This method works in O(1) get(key) { console.log(`Going to get the value for the key: ${key}`); return this .map.has(key) ? this .map.get(key) : -1; } // This method works in O(1) set(key, value) { console.log(`Going to set the (key, value): (${key}, ${value})`); if ( this .map.size >= this .CAPACITY) { // Remove the oldest entry if capacity is reached const oldestKey = this .map.keys().next().value; this .map. delete (oldestKey); } this .map.set(key, value); } } function main() { console.log( "Going to test the LRU Cache Implementation" ); const cache = new LRUCache(2); // it will store a key (1) with value 10 in the cache. cache.set(1, 10); // it will store a key (1) with value 10 in the cache. cache.set(2, 20); console.log(`Value for the key: 1 is ${cache.get(1)}`); // returns 10 // evicts key 2 and store a key (3) with value 30 in the cache. cache.set(3, 30); console.log(`Value for the key: 2 is ${cache.get(2)}`); // returns -1 (not found) // evicts key 1 and store a key (4) with value 40 in the cache. cache.set(4, 40); console.log(`Value for the key: 1 is ${cache.get(1)}`); // returns -1 (not found) console.log(`Value for the key: 3 is ${cache.get(3)}`); // returns 30 console.log(`Value for the key: 4 is ${cache.get(4)}`); // return 40 } // Execute the main function main(); |
Going to test the LRU Cache Implementation Going to set the (key, value) : (1, 10) Going to set the (key, value) : (2, 20) Going to get the value for the key : 1 Value for the key: 1 is 10 Going to set the (key, value) : (3, 30) Going to get the value for the key : 2 Value for the key: 2 is -1 Going to set the (key, value) : (4, 40) Going to get the value for the key : 1 Value for the key: 1 is -1 Going to get the value for the key : 3 Value for the key: 3 is 30 Going to get the value for the key : 4 Value for the key: 4 is 40
You can find the C++ code here
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