August, 2015 | Ajinkya Wavare

Simple Array Sum

Problem Statement

You are given an array of integers of size N. You need to print the sum of the elements of the array.

Input Format
The first line of the input consists of an integer N. The next line contains N space-separated integers describing the array.

Constraints
1≤N≤1000
0≤A[i]≤1000

Output Format
Output a single value equal to the sum of the elements of the array.

Sample Input

6
1 2 3 4 10 11

Sample Output

Explanation
1+2+3+4+10+11=31

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#define max 1000
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int n,temp,sum;
    cin>>n;
    while(n--){
        cin>>temp;
        sum+=temp;
    }
        cout<<sum;
    return 0;
}

Solve me second

Problem Statement

You learnt about STDIN and STDOUT in Solve me first.

This is the second challenge in the introduction series. The purpose of this challenge is to give you a working I/O template in your preferred language. It includes scanning two space-separated integers from STDIN in a loop over T lines, calling a function, returning a value, and printing it to STDOUT.

A pseudo code looks like the following:

read T
loop from 1 to T
    read A and B
    compute the sum
    print value in a newline
end loop

The task is to scan two numbers from STDIN, and print the sum A+B on STDOUT. The code has already been provided for most of the popular languages. This is primarily for you to read and inspect how the IO is handled.

Note: The code has been saved in a template, which you can submit if you want. Or, you may try rewriting it and building it up from scratch.

Input Format
(This section specifies the Input Format.)
The first line contains T (number of test cases) followed by T lines
Each line contains A and B, separated by a space.

As you can see that we have provided in advance the number of lines, we discourage the use of scanning till EOF as not every language has an easy way to handle that. In fact, every HackerRank challenge is designed in such a way that multitests begin with a T line to indicate the number of lines.

Output Format
(This section specifies the Output Format.)
An integer that denotes Sum (A+B) printed on new line for every testcase.

Constraints
(This section tells what input you can expect. You can freely assume that the input will remain within the boundaries specified.)
1≤T,A,B≤1000

Sample Input

2
2 3
3 7

Sample Output

5
10

The above sample should be taken seriously. 2 in the first line describes how many lines will follow, and your test cases are 2, 3 and 3, 7 in two separate lines. Your output should be 5 and 10 printed on two separate lines. If you print extra lines or “The answer is: 5”, any such extra characters in output will result in a Wrong Answer, as the judging is done using diff checker.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int solveMeSecond(int a, int b) {
  return a+b;
}

int main() {
  int num1, num2;
  int sum;
  int t;
  cin >> t;

  for (int i=0; i> num1 >> num2;
    sum = solveMeSecond(num1,num2);
    cout << sum << endl;
  }

  return 0;
}

Solve me first

Problem Statement

This is an introductory challenge. The purpose of this challenge is to give you a working I/O template in your preferred language. It includes scanning two integers from STDIN, calling a function, returning a value, and printing it to STDOUT.

Note: The code has been saved in a template, which you can submit if you want. Or, you may try rewriting it and building it up from scratch.

Input Format
(This section specifies the Input Format.)
Given A and B on two different lines.

Output Format
(This section specifies the Output Format.)
An integer that denotes Sum (A+B)

Constraints
(This section tells what input you can expect. You can freely assume that the input will remain within the boundaries specified. As an example here given below, A and B will never be below 1 or above 1000.)
1≤A,B≤1000

Sample Input

2
3

Sample Output

The above sample should be taken seriously. The input will be 2 and 3 in two separate lines, and the output should be just one number, 5. You should not print any whitespace at the beginning of output (e.g. ” 5″ or “\n5”), unless specifically asked for. Also, printing any extra non-whitespace characters such as “The answer is: 5” will result in a Wrong Answer, as the judging is done using diff checker.

Solution

  
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {


    static int solveMeFirst(int a, int b) {
        return a+b;
   }

   
 public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int _a;
        _a = in.nextInt();
        int _b;
        _b = in.nextInt();
        int sum;
        sum = solveMeFirst(_a, _b);
        System.out.println(sum);
   }
}

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

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;
    
    }
        
}

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