## Sunday, 11 October 2015

## Wednesday, 23 September 2015

# Hackerrank Insert a node into a sorted doubly linked list Solution

##
**Problem Statement**

You’re given the pointer to the head node of a sorted doubly linked list and an integer to insert into the list. Create a node and insert it into the appropriate position in the list. The head node might be NULL to indicate that the list is empty.##
**Source Code:**

```
Node* SortedInsert(Node *head,int data)
{
// Complete this function
// Do not write the main method.
Node *temp = new Node;
temp->data = data;
temp->next = NULL;
temp->prev = NULL;
if(head == NULL){
head = temp;
}else{
Node *temp1 = temp;
temp->next = head;
head = temp;
while(temp!= NULL){
temp1 = temp;
temp = temp->next;
if(temp!=NULL){
if(temp1->data>temp->data){
int tdata;
tdata = temp1->data;
temp1->data = temp->data;
temp->data = tdata;
}
}
}
}
return head;
}
```

** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem but it passes all the testcases in Hackerrank. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)

## Thursday, 6 August 2015

# GRK Web

We are a passionate Computer Science Engineering Students who believe in our dreams.

We are application Developers who develop awesome applications for PC and Mobile platforms.

We build the Future Today.

Our Official website : grkweb.com

Facebook Page : facebook.com/grkweb

## Wednesday, 5 August 2015

# C++ Diamond Pattern

C++ program to print the diamond pattern using '#' symbols. In this program two for loops are used to print the pattern. The first loop prints the upper part to the diamond ( triangle part ). The second for loop prints the lower part of the diamond (an inverted triangle ) . By combining both of the triangles an diamond pattern structure is formed.

## PROGRAM CODE:

```
#include<iostream>
using namespace std;
int main()
{
int i,j,k=1,s=6,l;
for(i=0;i<5;i++,k-=2,s-=1)
{
for(l=s-1;l>=0;l--)
cout<<' ';
for(j=k;j<0;j++)
cout<<"#";
cout<<endl;
}
k = 9;
s = 0;
for(i=0;i<5;i++,k-=2,s+=1)
{
for(l=s;l>=0;l--)
cout<<' ';
for(j=k;j>0;j--)
cout<<"#";
cout<<endl;
}
return 0;
}
```

## OUTPUT:

#

# # #

# # # # #

# # # # # # #

# # # # # # # # #

# # #

# # # # #

# # # # # # #

# # # # # # # # #

# # # # # # #

# # # # #

# # #

# # # # #

# # #

#

** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)

## Saturday, 18 April 2015

# Hackerrank Print the elements of a linked list solution

## 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!

## Source code:

```
/*
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;
}
}
```

** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem but it passes all the testcases in Hackerrank. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)

# Hackerrank Print in Reverse solution

## Problem Statement

This challenge is part of a tutorial track by MyCodeSchool and is accompanied by a video lesson.YouĂ˘€™re 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, donĂ˘€™t print anything!

## Soruce code:

```
/*
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.
int a[100],i=0;
while(head!=NULL){
a[i] = head->data;
i++;
head = head->next;
}
for(int j = i-1;j>=0;j--){
cout<<a[j]<<endl;
}
}
```

** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem but it passes all the testcases in Hackerrank. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)# Hackerrank Reverse a doubly linked list solution

## Problem Statement

This challenge is part of a tutorial track by MyCodeSchoolYou’re given the pointer to the head node of a doubly linked list. Reverse the order of the nodes in the list. The head node might be NULL to indicate that the list is empty.

## Source Code:

** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem but it passes all the testcases in Hackerrank. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)`/* Reverse a doubly linked list, input list may also be empty Node is defined as struct Node { int data; Node *next; Node *prev; } */ Node* Reverse(Node* head) { Node *cur = head,*temp = new Node; // Complete this function // Do not write the main method. while(cur !=NULL){ temp->next = cur->next; temp->prev = cur->prev; cur->next = temp->prev; cur->prev = temp->next; cur = temp->next; if(cur!=NULL){ head = cur; } } return head; }`

# Hackerrank Merge two sorted linked lists solution

## Problem Statement

This challenge is part of a tutorial track by MyCodeSchool

You’re given the pointer to the head nodes of two sorted linked lists. The data in both lists will be sorted in ascending order. Change the next pointers to obtain a single, merged linked list which also has data in ascending order. Either head pointer given may be null meaning that the corresponding list is empty.

## Source code:

** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem but it passes all the testcases in Hackerrank. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)`/* Merge two sorted lists A and B as one linked list Node is defined as struct Node { int data; struct Node *next; } */ Node* MergeLists(Node *headA, Node* headB) { // This is a "method-only" submission. // You only need to complete this method Node *head=NULL, *cur,*prev = NULL; Node *curA,*curB; head = headA; cur = head; curB = headB; while(curB !=NULL){ if(cur== NULL){ head = headB; break; }else{ if(cur->data > curB->data){ curA = curB; curB = curB->next; curA->next = cur; if(prev == NULL){ head = curA; }else{ prev->next = curA; } prev = cur; cur = cur->next; }else{ if(cur->next !=NULL){ prev = cur; cur = cur->next; }else{ cur->next = curB; break; } }} }return head; }`

# Hackerrank Reverse a linked list solution

## 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.

## source code:

```
/*
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 = NULL;
Node *cur = head;
Node *next ;
while(cur!=NULL){
next = cur->next;
cur->next = prev;
prev = cur;
cur = next;
}
head = prev;
return head;
}
```

** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem but it passes all the testcases in Hackerrank. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)# Hackerrank Insert a node at a specific position in a linked list solution

## 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.

## source code:

```
/*
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 *cur = new Node,*temp = new Node;
temp->data = data;
temp->next = NULL;
if(head == NULL){
head = temp;
}else{
int i=0;
cur = head;
while(cur!=NULL){
if(position == 0){
temp->next = cur;
head= temp;
}
else if(i==position-1){
temp->next = cur->next;
cur ->next = temp;
}else{
}i++;
cur= cur->next;
}
}
return head;
}
```

** The above solution is my own code and it may not be the optimal solution or optimal way to approach the problem but it passes all the testcases in Hackerrank. So if you have any optimal approaches feel free to paste the code as the comment below..... :) :) :)