LRU cache | Live Coding with Explanation | Leetcode

[@AlgorithmsMadeEasy]

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[@shivendra2076]

Under rated channel, Thanks bro, understood it

[@AlgorithmsMadeEasy]

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[@pranavananth6617]

one of the best solution...

[@AlgorithmsMadeEasy]

Thank you!!

[@dambar67]

map.remove(node), how it will remove ,here key should be given not the value.

[@appprocessorsappdevelopmen8405]

*insert() method without declaring headNext*
void insert(Node node){
//insert into map
map.put(node.key,node);
//point node's next to head's next
node.next = head.next;
//point head's next' prev to node
head.next.prev = node;
head.next=node;
node.prev=head;
}

[@milenitrivedi7561]

Great explanation !

[@RupamSasmalYt]

Have done the same approach with c++, but it's giving TLE. Can't find any single solution in c++ without tle in leetcode :(
C++ Code:
class Node{
public:
int key, val;
Node *prev, *next;
Node(int key,int val){
this->key=key;
this->val=val;
}
};
class LRUCache {
public:
int cap;
Node* head=new Node(0,0);
Node* tail=new Node(0,0);
unordered_map mp;

LRUCache(int capacity) {
cap=capacity;
head->next=tail;
tail->prev=head;
}
void Add(Node* newNode){
mp[newNode->key]=newNode;
Node* tmp=head->next;
head->next=newNode;
newNode->prev=head;
newNode->next=tmp;
tmp->prev=newNode;
}
void Delete(Node* node){
mp.erase(node->key);
node->prev->next=node->next;
node->next->prev=node->prev;
}
int get(int key) {
if(mp.find(key)!=mp.end()){ // key present
Node* resNode=mp[key];
Delete(resNode);
Add(resNode);
return resNode->val;
}
return -1;
}

void put(int key, int value) {
if(mp.find(key)!=mp.end()){ // key is present
Delete(mp[key]);
}
if(mp.size()==cap){
Delete(tail->prev);
}
Add(new Node(key,value));
}
};

[@omarkhan5223]

@5:40 You remove tail.prev. I thought you would remove tail because the tail of linked list is the last node, isn't tail.prev the second last node?? Why is this not the case?

[@054_heenaahmed8]

tail is the dummy node that is at end of linkedlist , similar to head , which is also dummy at start of linkedlist . we have used it to simplify the operations , and get access to last(using tail) and first(using head) node in O(1).

[@rilkedev449]

@@054_heenaahmed8 Thank you.

[@sayyidiskandarkhan3064]

it would have been great had he covered this in the explaination but i had to look at the comments to find the results.

[@CRamPrasannaCV]

Understood :-)