Tag: Linked Lists

Insert a node at the head of a linked list

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.

You’re given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer, insert this node at the head of the linked list and return the new head node. The head pointer given may be null meaning that the initial list is empty.

Input Format
You have to complete the Node* Insert(Node* head, int data) method which takes two arguments – the head of the linked list and the integer to insert. You should NOT read any input from stdin/console.

Output Format
Insert the new node at the head and return the head of the updated linked list. Do NOT print anything to stdout/console.

Sample Input

NULL , data = 1
1 –> NULL , data = 2

Sample Output

1 --> NULL
2 --> 1 --> NULL

Explanation
1. We have an empty list, on inserting 1, 1 becomes new head.
2. We have a list with 1 as head, on inserting 2, 2 becomes the new head.

Solution

  
/*
  Insert Node at the begining of a linked list
  Initially head pointer argument could be NULL for empty list
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
return back the pointer to the head of the linked list in the below method.
*/
Node* Insert(Node *head,int data)
{
  // Complete this method
    Node *newnode = (Node *)malloc(sizeof(Node));
    newnode->data=data;
    if(head==NULL){
        head=newnode;
        newnode->next=NULL;
    }
    else{
        newnode->next=head;
        head=newnode;
    }
    return head;
        
}

Get Node Value

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool

You’re given the pointer to the head node of a linked list and a specific position. Counting backwards from the tail node of the linked list, get the value of the node at the given position. A position of 0 corresponds to the tail, 1 corresponds to the node before the tail and so on.

Input Format
You have to complete the int GetNode(Node* head, int positionFromTail) method which takes two arguments – the head of the linked list and the position of the node from the tail. positionFromTail will be at least 0 and less than the number of nodes in the list. You should NOT read any input from stdin/console.

Constraints
Position will be a valid element in linked list.

Output Format
Find the node at the given position counting backwards from the tail. Then return the data contained in this node. Do NOT print anything to stdout/console.

Sample Input

1 -> 3 -> 5 -> 6 -> NULL, positionFromTail = 0
1 -> 3 -> 5 -> 6 -> NULL, positionFromTail = 2

Sample Output

6
3

Solution

  
/*
  Get Nth element from the end in a linked list of integers
  Number of elements in the list will always be greater than N.
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
int GetNode(Node *head,int positionFromTail)
{
  // This is a "method-only" submission. 
  // You only need to complete this method.
    Node *ptr1=head;
    Node *ptr2=head;
    while(positionFromTail--){
        ptr2=ptr2->next;
    }
    while(ptr2->next!=NULL){
        ptr1=ptr1->next;
        ptr2=ptr2->next;
    }
    return ptr1->data;
}

Compare two linked lists

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. The lists are equal only if they have the same number of nodes and corresponding nodes contain the same data. Either head pointer given may be null meaning that the corresponding list is empty.

Input Format
You have to complete the int CompareLists(Node* headA, Node* headB) method which takes two arguments – the heads of the two linked lists to compare. You should NOT read any input from stdin/console.

Output Format
Compare the two linked lists and return 1 if the lists are equal. Otherwise, return 0. Do NOT print anything to stdout/console.

Sample Input

NULL, 1 –> NULL
1 –> 2 –> NULL, 1 –> 2 –> NULL

Sample Output

0
1

Explanation
1. We compare an empty list with a list containing 1. They don’t match, hence return 0.
2. We have 2 similar lists. Hence return 1.

Solution

  
/*
  Compare two linked lists A and B
  Return 1 if they are identical and 0 if they are not. 
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
int CompareLists(Node *headA, Node* headB)
{
  // This is a "method-only" submission. 
  // You only need to complete this method
 /*   if(headA==NULL && headB==NULL)
        return 1;
     else if(headA!=NULL || headB==NULL)
        return 0;
     else {
     if(headA!=NULL && headB!=NULL)
    {
        if(headA->data == headB->data)
            CompareLists(headA->next,headB->next);
        else return 0;    
    }
     if(headA==NULL && headB==NULL)
                return 1;
            else
                return 0;
     }
    
        
    return 0;*/
    
        if(headA==NULL || headB==NULL)
            return 0;
           else
            {
            while(headA!=NULL && headB!=NULL)
            {
                if(headA->data==headB->data)
               { headA=headA->next;
                headB=headB->next;}
                else 
                    return 0;
            }
            if(headA==NULL && headB==NULL)
                return 1;
            else
                return 0;
        }
}

Reverse a linked list

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.

You’re given the pointer to the head node of a linked list. Change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty.

Input Format
You have to complete the Node* Reverse(Node* head) method which takes one argument – the head of the linked list. You should NOT read any input from stdin/console.

Output Format
Change the next pointers of the nodes that their order is reversed and return the head of the reversed linked list. Do NOT print anything to stdout/console.

Sample Input

NULL
2 –> 3 –> NULL

Sample Output

NULL
3 --> 2 --> NULL

Explanation
1. Empty list remains empty
2. List is reversed from 2,3 to 3,2

Solution

  
/*
  Reverse a linked list and return pointer to the head
  The input list will have at least one element  
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
Node* Reverse(Node *head)
{
  // Complete this method
    Node *prev,*current,*forward;
    current=head;
    prev=NULL;
    while(current!=NULL){
        forward=current->next;
        current->next=prev;
        prev=current;
        current=forward;
        
    }
    head=prev;
    return head;
}

Print in Reverse

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.

You are given the pointer to the head node of a linked list and you need to print all its elements in reverse order from tail to head, one element per line. The head pointer may be null meaning that the list is empty – in that case, do not print anything!

Input Format
You have to complete the void ReversePrint(Node* head) method which takes one argument – the head of the linked list. You should NOT read any input from stdin/console.

Output Format
Print the elements of the linked list in reverse order to stdout/console (using printf or cout) , one per line.

Sample Input

1 –> 2 –> NULL
2 –> 1 –> 4 –> 5 –> NULL

Sample Output

2
1
5
4
1
2

Explanation
1. First list is printed from tail to head hence 2,1
2. Similarly second list is also printed from tail to head.

Solution

  
/*
  Print elements of a linked list in reverse order as standard output
  head pointer could be NULL as well for empty list
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
void ReversePrint(Node *head)
{
  // This is a "method-only" submission. 
  // You only need to complete this method. 
    // Base case  
  if(head == NULL)
    return;
 
  // print the list after head node
  ReversePrint(head->next);
 
  // After everything else is printed, print head
  cout<<head->data<<endl;
}

Delete a Node

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.

You’re given the pointer to the head node of a linked list and the position of a node to delete. Delete the node at the given position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The list may become empty after you delete the node.

Input Format
You have to complete the Node* Delete(Node* head, int position) method which takes two arguments – the head of the linked list and the position of the node to delete. You should NOT read any input from stdin/console. position will always be at least 0 and less than the number of the elements in the list.

Output Format
Delete the node at the given position and return the head of the updated linked list. Do NOT print anything to stdout/console.

Sample Input

1 –> 2 –> 3 –> NULL, position = 0
1 –> NULL , position = 0

Sample Output

2 --> 3 --> NULL
NULL

Explanation
1. 0th position is removed, 1 is deleted from the list.
2. Again 0th position is deleted and we are left with empty list.

Solution

  
/*
  Delete Node at a given position in a linked list 
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
Node* Delete(Node *head, int position)
{
  // Complete this method
    Node *  temp = head, *prev;
    if(position == 0){
        temp= head->next;
        free(head);
        return temp;
    }
    else{
        while(position--){
            prev=temp;
            temp=temp->next;
        }
        prev->next=temp->next;
        free (temp);
        return head;
        
    }
}

Insert a node at a specific position in a linked list

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.

You’re given the pointer to the head node of a linked list, an integer to add to the list and the position at which the integer must be inserted. Create a new node with the given integer, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is empty.

Input Format
You have to complete the Node* Insert(Node* head, int data, int position) method which takes three arguments – the head of the linked list, the integer to insert and the position at which the integer must be inserted. You should NOT read any input from stdin/console. position will always be between 0 and the number of the elements in the list (inclusive).

Output Format
Insert the new node at the desired position and return the head of the updated linked list. Do NOT print anything to stdout/console.

Sample Input

NULL, data = 3, position = 0
3 –> NULL, data = 4, position = 0

Sample Output

3 --> NULL
4 --> 3 --> NULL

Explanation
1. we have an empty list and position 0. 3 becomes head.
2. 4 is added to position 0, hence 4 becomes head.

Note
For the purpose of evaluation the list has been initialised with a node with data=2. Ignore it, this is done to avoid printing empty lists while comparing output.

Solution


/*
  Insert Node at a given position in a linked list 
  The linked list will not be empty and position will always be valid
  First element in the linked list is at position 0
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
Node* InsertNth(Node *head, int data, int position)
{
  // Complete this method only
  // Do not write main function.
    Node * newnode = (Node *)malloc(sizeof(Node));
    newnode->data=data;
    Node *temp=head,*prev;
    if(position==0){
        newnode->next=head;
        head=newnode;
        return head;
    }
    else{
        while(position--){
            prev=temp;
            temp=temp->next;
        }
        prev->next=newnode;
        newnode->next=temp;
        return head;
    
    }
        
}

Insert a node at the tail of a linked list

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.

You’re given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer, insert this node at the tail of the linked list and return the head node. The head pointer given may be null meaning that the initial list is empty.

Input Format
You have to complete the Node* Insert(Node* head, int data) method which takes two arguments – the head of the linked list and the integer to insert. You should NOT read any input from stdin/console.

Output Format
Insert the new node at the tail and just return the head of the updated linked list. Do NOT print anything to stdout/console.

Sample Input

NULL, data = 2
2 –> NULL, data = 3

Sample Output

2 -->NULL
2 --> 3 --> NULL

Explanation
1. We have an empty list and we insert 2.
2. We have 2 in the tail, when 3 is inserted 3 becomes the tail.

Solution


/*
  Insert Node at the end of a linked list 
  head pointer input could be NULL as well for empty list
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
Node* Insert(Node *head,int data)
{
  // Complete this method
    Node *temp = head;
    Node *newnode = (Node *)malloc(sizeof(Node));
    newnode->data = data;
    newnode->next = NULL;
    if(temp==NULL){
        head= newnode;
        return head;
    }
    else {
        while (temp->next!= NULL)
            temp = temp->next;
        temp->next=newnode;
        return head;
    }   
}

Print the elements of a linked list

Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.

If you’re new to working with linked lists, this is a great exercise to get familiar with them. You’re given the pointer to the head node of a linked list and you need to print all its elements in order, one element per line. The head pointer may be null, i.e., it may be an empty list. In that case, don’t print anything!

Input Format
You have to complete the void Print(Node* head) method which takes one argument – the head of the linked list. You should NOT read any input from stdin/console. There are multiple test cases. For each test case, this method will be called individually.

Output Format
Print the elements of the linked list to stdout/console (using printf or cout) , one per line.

Sample Input

NULL  
1->2->3->NULL

Sample Output

1
2
3

Explanation

For first case, an empty list is passed to the method. So nothing is printed. For second case, all the elements of the linked list (1, 2 and 3) are printed in separate lines.

 
Solution


/*
  Print elements of a linked list on console 
  head pointer input could be NULL as well for empty list
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
void Print(Node *head)
{
  // This is a "method-only" submission. 
  // You only need to complete this method.
            while(head!= NULL){
            cout<< head->data<<endl;
                head=head->next;
            }
        
}