June 7, 2014

Just finding things in internet (ref provided) and solving them in JavaScript

- Stack and Queue in JavaScript
- Priority Queue
- Singly Linked List
- Doubly Linked List
- Remove a node from Singly Linked List
- Reverse a Singly Linked List
- Reverse a Doubly Linked List
- Find Kth Node from End
- Delete Kth Node from End
- Delete a node from Doubly Linked List
- Detect loop in a Singly Linked List
- Detect the node where loop started
- Find the middle node of a Singly LL
- Find length of a Singly Linked List
- Check duplicate value in a Singly with loop
- Rotate a Linked list by Kth Node
- Insert node into a sorted LL
- Merge two unsorted Linked Lists
- Find intersect of two Linked List
- sort Linked List
- remove duplicate from unsorted LL
- Create a Circular Linked List
- Check Palindrome in Linked List
- Deleted Questions

**Question:** How will you implement stack and queue with JavaScript?

**Answer: **stack is already implemented in JavaScritp array. you can simply call `push`

and `pop`

method.

```
var myStack = [];
//push
myStack.push(1);
myStack.push(2);
myStack.push(3);
//pop
myStack.pop(); //3
myStack.pop(); //2
myStack.pop(); //1
```

Queue is pretty much same. other than calling `pop`

you will use `shift`

method to get the element in the front side of your array.

```
var myQueue = [];
//push
myQueue.push(1);
myQueue.push(2);
myQueue.push(3);
//pop
myQueue.shift(); //1
myQueue.shift(); //2
myQueue.shift(); //3
```

you have to use a maximizing heap to manage a priority queue. Hence while extracting, you just get the one on the top and while inserting you will do it at the bottom

```
//for now read the link.
```

ref: i have implemented simplified version of priorityQueueJS

**Question: ** How will you create a linked list in JavaScript?

**Answer: **if you don't have any idea about linked list. read: wiki: linked list

to have linked list, you need a node object that will point to the next. you need one pointers (head) to point the first node of your linked list. Besides, you need methods to add and remove node from your linked list

let's get started. We have an object LinkedList and that has one property head (a pointer)

```
function LinkedList(){
this.head = null;
}
```

to create add elements, will use a push method in the prototype of the object. push method will take value and will create a node object. if there is no head, then node will be the value of head. if there is a head. then we have to walk through the linked list chain to read the tail (tail is where the next node is null). We will simple set the of next at the tail to node.

```
LinkedList.prototype.push = function(val){
var node = {
value: val,
next: null
}
if(!this.head){
this.head = node;
}
else{
current = this.head;
while(current.next){
current = current.next;
}
current.next = node;
}
}
```

Now. create a linked list and push values in it.

```
var sll = new LinkedList();
//push node
sll.push(2);
sll.push(3);
sll.push(4);
//check values by traversing LinkedList
sll.head; //Object {data: 2, next: Object}
sll.head.next; //Object {data: 3, next: Object}
sll.head.next.next; //Object {data: 4, next: null}
```

read: Advantage and disadvantage of linked list

copied from: Nicholas Zakas: linked list

for doubly linked list there will be a link backward to the previous element. Your node object will look like

```
var node = {
value: val,
next: null,
previous: null
}
```

```
function DoublyLinkedList(){
this.head = null;
}
DoublyLinkedList.prototype.push = function(val){
var head = this.head,
current = head,
previous = head;
if(!head){
this.head = {value: val, previous:null, next:null };
}
else{
while(current && current.next){
previous = current;
current = current.next;
}
current.next = {value: val, previous:current, next:null}
}
}
```

```
//test at least once
var dll = new DoublyLinkedList();
dll.push(2);
dll.push(3);
dll.push(4);
dll.push(5);
//trust me it works
```

get most of it from: doubly linked list

**Question:** How could you write an extension to remove a node from a LL

If you want to remove a node from your linked list you have to find the node. There are three conditions here

- case -1: your targeted node is in the head. you have to replace the head with the next node
- case -2: your targeted node is in the tail. you just have to remove it from the tail. Hence next of the node before the tail will be null.
- case-3: your targeted node is in the middle of the LinkedList, you have to make the node after your targeted node to be the next node of the node before your targeted node.

let's do it

```
LinkedList.prototype.remove = function(val){
var current = this.head;
//case-1
if(current.value == val){
this.head = current.next;
}
else{
var previous = current;
while(current.next){
//case-3
if(current.value == val){
previous.next = current.next;
break;
}
previous = current;
current = current.next;
}
//case -2
if(current.value == val){
previous.next == null;
}
}
}
```

Now if you want to remove you can just pass the value

```
sll.remove(3);
sll.remove(1);
```

**Question: ** How would you reverse a singly LL (no loop)?

**Answer: ** Just walk through the LL and put the nodes in an array. This would be easier other than using remove function (if LL have one), because to remove a single item from the end of the SLL you have to walk all way to the end of the SLL every single time. Here you are just walking once.

```
function reversesll(sll){
if(!sll.head || !sll.head.next) return sll;
var nodes=[],
current = sll.head;
//storing all the nodes in an array
while(current){
nodes.push(current);
current = current.next;
}
var reversedLL = new LinkedList();
reversedLL.head = nodes.pop();
current = reversedLL.head;
var node = nodes.pop();
//make sure to make next of the newly inserted node to be null
//other wise the last node of your SLL will retain its old next.
while(node){
node.next = null;
current.next = node;
current = current.next;
node = nodes.pop();
}
return reversedLL;
}
```

test at least once.

```
//create the LL
var sll = new LinkedList();
sll.push(1);
sll.push(2);
sll.push(3);
sll.push(4);
sll.push(5);
//test it
reversesll(sll);
//{head: {value:5, next:{value: 4, next: {value: 3, next: {value:2, next:{value:1, next: null}}}}}}
```

```
//search it and implement it
```

**question**Given number k, for Singly linked list, skip k nodes and then reverse k nodes, till the end.

**Question:** reverse the doubly linked list without using extra space

**Answer:**

```
function reverseDoublyLL(dll){
var head = dll.head,
current = dll.head,
tmp;
while(current){
tmp = current.next;
current.next = current.previous;
current.previous = tmp;
if(!tmp){
//set the last node as header
dll.head = current;
}
current = tmp;
}
return dll;
}
```

```
//test it at least once
//or trust me it works
```

ref: career cup

**Question: ** How could you get the Kth node from the end (no loop)

**Answer: **

```
function kthFromEnd(sll, k){
var node = sll.head,
i = 1,
kthNode;
//handle, 0 or negative value of k
if(k<=0) return;
while(node){
if(i == k) kthNode = sll.head;
else if(i-k>0){
kthNode = kthNode.next;
}
i++;
node = node.next;
}
return kthNode;
}
```

```
//create the LL
var sll = new LinkedList();
sll.push(1);
sll.push(2);
sll.push(3);
sll.push(4);
sll.push(5);
//test at least once
kthFromEnd(sll, 1); //Object {value: 5, next: null}
kthFromEnd(sll, 3); //Object {value: 3, next: Object}
```

**Question: ** How could you delete kth node from the end of a singly LL (no loop)

```
function deleteKthFromEnd(sll, k){
var node = sll.head,
i = 1,
kthNode,
previous;
if(k<=0) return sll;
while(node){
if(i == k) kthNode = sll.head;
else if(i-k>0){
previous = kthNode;
kthNode = kthNode.next;
}
i++;
node = node.next;
}
//kth node is the head
if(!previous)
sll.head = sll.head.next;
else{
previous.next = kthNode.next;
}
return sll;
}
```

test at least once

```
//test at least once
var sll = new LinkedList();
sll.push(1);
sll.push(2);
sll.push(3);
sll.push(4);
sll.push(5);
deleteKthFromEnd(sll, 2);
//{head: {value:1, next:{value: 2, next: {value: 3, next: {value:1, next:null}}}}}
```

**Question:** How could you delete a node from a doubly linked list?

**Answer: ** there could be four cases

**delete head:** very simple, make the second node as head. and dont forget to set previous of new head to null.

**node in the middle: ** skip break the chain, skip current node. hence next of previous node would be the next of the current node. Besides, previous of the next node would be previous node

**delete tail: ** its also simple, make the next of previous as null.

**didn't find:** dont do anything

```
function deleteNodeDLL(val){
var current = dll.head,
previous;
//delete head
if(current.value == val){
dll.head = current.next;
//if there is only one node, then dll.head is null
if(dll.head) dll.head.previous = null;
return dll;
}
while(current.next){
if(current.value == val){
previous.next = current.next;
current.next.previous = previous;
return dll;
}
previous = current;
current = current.next;
}
//delete last node
if(current.value == val){
previous.next = null;
}
return dll;
}
```

```
function detectLoop(sll){
var slowPointer = sll.head,
fastPointer = sll.head;
while(slowPointer && fastPointer && fastPointer.next){
slowPointer = slowPointer.next;
fastPointer = fastPointer.next.next;
if(slowPointer == fastPointer){
return true;
}
}
return false;
}
```

```
//test at least once
var sll = new LinkedList();
sll.push(1);
sll.push(2);
sll.push(3);
sll.push(4);
sll.push(5);
detectLoop(sll); //false
//create a loop manually. point next of 5 to point 3
sll.head.next.next.next.next.next = sll.head.next.next
detectLoop(sll); //true
```

**Question:**

**Answer:** you will have two pointer slow and fast like you had to detect loop and they will meet somewhere. if fast one hit null (if thee is a way to hit null, fast one should hit if before the slower. because fast one moves fast (recally physics theories!)), there is no loop and returns null.

if there is a null and whenever both pointers meet, set slow pointer as head. Move both the pointer move by one step at a time

At the time of meet, fast one moved twice as much as slow one.

**think about it and find a way to explain it**

```
function findLoopStart(sll){
var slow = sll.head,
fast = sll.head;
while(slow && fast){
slow = slow.next;
//if hits null, then there is no loop
if(!fast.next){
return null;
}
fast = fast.next.next;
if(slow == fast){
slow = sll.head;
while(slow != fast){
slow = slow.next;
fast = fast.next;
}
return slow;
}
}
}
```

**Question:** How could you detect middle of a LL?

**Answer: ** need two pointer one will move two step another one. when two step will hit null. you will get middle for the singly. think about it.

```
//If you cant figure it out ourself. don't read further, do something else
```

**Question:** there could be loop in the singly LL

**Answer:**

```
//doesnt work.. somewhere gets stack overflow
//debug and find fix it
function getLength(sll){
var head = sll.head,
current = head,
pointer = head,
anotherPtr,
len = 0;
//use the previously written function
var loopStartNode = findLoopStart(sll);
if(!loopStartNode){
while(current){
current= current.next;
len++;
}
return len;
}
else{
anotherPtr = loopStartNode;
while(pointer != anotherPtr){
len += 2;
pointer = pointer.next;
anotherPtr = anotherPtr.next;
}
return len;
}
}
```

**Question:** Write a function that will return true if a circular singly linked list has duplicate values

**Answer:**

```
//code it man! dont be lazy
```

Similar question: Find the length of a linked list which contains cycle.

```
//just start a counter instead of checking duplicate
```

ref: career cup

**Question:** How could you rotate a Linked List by Kth Node?

or

**Question:**Append the last n nodes of a linked list to the beginning of the list

**Answer: ** if the given linked list is: 1->2->3->4->5 and k is 3,the list should be modified to: 4->5->1->2->3.

```
//will not work for k less than length-1
function rotateByKthNode(sll, k){
var prevHead = sll.head,
previous = sll.head,
current = sll.head,
i = 1;
while(current.next){
if(i==k+1){
sll.head = current;
previous.next = null;
}
previous = current;
current = current.next;
i++;
}
current.next = prevHead;
return sll;
}
```

```
rotateByKthNode(sll,3);
//{head: {value:4, next:{value: 5, next: {value: 1, next: {value:2, next:{value:3, next: null}}}}}}
```

ref: rotate by kth Node

**Question:** Insert node in a sorted LL

**Answer:**

```
function pushSorted(sll, val){
var head = sll.head,
current = head,
previous;
//value lower than head value
if(val < sll.head.value){
sll.head = {value: val, next: head};
return sll;
}
while(current){
if(current.value > val){
previous.next = {value: val, next: current};
return sll;
}
previous = current;
current = current.next;
}
//value higher than the last node value
previous.next = {value:val, next: null};
return sll;
}
```

```
var sll = new LinkedList();
sll.push(5);
sll.push(7);
sll.push(10);
sll.push(14);
pushSorted(sll, 9);
//{head: {value: 5, next: {value: 7, next: {value: 9, next: {value: 10, next: {value: 14, next: null}}}}}}
```

**Question:** How would you merge two unsorted linked list?

**Answer:**

ref: merge two unsorted linked list to one sorted LL

easier question: merge two LL in alternate position

**Question:** How could you find intersect of two SLL in single iteration?

**Answer:**

ref: most detail: have 6 methods or has some animation or intersection of two LL or career cup

**Question:** How would you sort a linked list?

**Answer:** you can use bubble, merge or similar sort

```
```

ref: sort LL

**Question:** Given an integer linked list of which both first half and second half are sorted independently. Write a function to merge the two parts to create one single sorted linked list in place [do not use any extra space].

**Answer:**

ref: career cup

**Question:** Delete the repeated elements in a unsorted singly linked list in O(n) time complexity without using extra space.

**Answer:**

```
//get a crystal clear concept about space complexity
//also this question is in cracking the coding interview
```

ref: career cup

If the next pointer of the last node points to the head node, it becomes circular linked list. On Singly or Doubly Linked List, you will know you hit the tail when you hit null. For circular linked list, you will know you hit the tail when you hit the head (the next of last one pointed to head).

```
function CircularLinkedList(){
this.head = null;
}
CircularLinkedList.prototype.push = function(val){
var head = this.head,
current = head,
node = {value: val, next: null, previous: null};
if(!head){
node.next = node;
node.previous = node;
this.head = node;
}
else{
while(current.next != head){
current = current.next;
}
node.next = head;
node.previous = current;
head.previous = node;
current.next = node;
}
}
```

```
var cll = new CircularLinkedList();
cll.push(3);
cll.push(4);
cll.push(5);
//trust me dude, it works
```

Now you can play with it to remove, sort, or insert into sorted or remove duplicate, blah blah blah

ref: circular linked list

**Question:**

**Answer:**

```
//get the solution from cracking the coding interview
//or ggl it
```

- implement queue by using stack
- queue implement with two stack
- implement circular queue
- frequencey of given number in LL
- delete alternative nodes
- move last node to the front
- youtube have stuff as well
- Given a binary tree, convert it into a doubly circular linked list. or ternary tree to LL
- Delete a node from SLL, in which the last node points to the middle node( in case of even no of nodes, it points to the first middle node) and update the SLL.
- how to sort a single linked list with out using an additional node?
- swap the consecutive elements of the list and return the head node. or swap every two nodes
- Given a doubly linked list containing only three integers 1,2,3. Sort the list without exchanging the values.
- You are given two numbers in the form of linked list.Add them without reversing the linked lists
- Write a program to reverse every K elements of a linked list.
- Input: 1->2->3->4->a->b->c->d->5->6->e->f Output:1->a->2->b->3->c->4->d->5->e->6->f
- Given a linked list with next and high pointers, populate high pointers to the next higher node, inplace and O(n).
- Insert node to a sorted circular Linked List

Feel free to express your anger (sorry folks, you have to use g+.). Also point out my mistakes ( technical, wrong answer, spelling, grammar, sentence..., whatever), let your dude learn and grow.