JS: Interview Questions

Basic Concept (beginners level)

1. Stack and Queue in JavaScript
2. Priority Queue
3. Singly Linked List
4. Doubly Linked List
5. Remove a node from Singly Linked List
6. Reverse a Singly Linked List
7. Reverse a Doubly Linked List
8. Find Kth Node from End
9. Delete Kth Node from End
10. Delete a node from Doubly Linked List
11. Detect loop in a Singly Linked List
12. Detect the node where loop started
13. Find the middle node of a Singly LL
14. Find length of a Singly Linked List
15. Check duplicate value in a Singly with loop
16. Rotate a Linked list by Kth Node
17. Insert node into a sorted LL
18. Merge two unsorted Linked Lists
19. Find intersect of two Linked List
20. sort Linked List
21. remove duplicate from unsorted LL
22. Create a Circular Linked List
23. Check Palindrome in Linked List
24. Deleted Questions

Stack and Queue

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

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
        

Priority Queue

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

Singly Linked List

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

Doubly 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

Remove Node from Singly LL

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); 
        

remove last element doesnt work

Reverse Singly Linked List

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.

Iterative

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}}}}}}
        

Recursive

//search it and implement it
        

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

Reverse Doubly LL

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

kth node from End

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}
        

Delete kth node from the end

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}}}}}
        

Delete a node from doubly LL

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;
}
        

Detect a loop in Singly Linked List

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
        

Detect Node where loop started

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;
       }
   }
}
        

ref: detect where loop started

Find middle

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
        

Length of a Singly LL

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;
    }
}
        

Loop with duplicate

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

Rotate by Kth Node

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

Insert Node to Sorted LL

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}}}}}}
        

merge two unsorted array

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

Intersect of two LL

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

sort Linked list

Question: How would you sort a linked list?

Answer: you can use bubble, merge or similar sort

ref: sort LL

toggle as hard questions

special sort

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

Remove duplicate from unsorted LL

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

circular Linked List

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

Check whether a linked list is a palindrome

Question:

Answer:

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

Deleted Questions

• 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

Express anger!

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.