Linked List Insert nth problem

A linked list problem for inserting a node after the nth index !!

InsertNth() Solution


void InsertNth(struct node** headRef, int index, int data)
if (index == 0)
   Push(headRef, data);                       // position 0 is a special case...
else {
  struct node* current = *headRef;
   int i;
     for (i=0; i<index-1; i++) {
     assert(current != NULL); // if this fails, index was too big
     current = current->next;
      }
Push(&(current->next), data);
}
}

Problem on Bit Manipulation

Write a function to determine the number of bits required to convert integer X to integer Y.
Input: 31, 14
Output: 2


This problem seems to be quite complex, however is actually rather simple. Think about finding out which bits in two numbers are different. Simple: with an XOR.
Each 1 in the XOR will represent one different bit between X and Y. Then we can simply count the number of bits that are 1.

 public static int swapRequired(int x, int y)
{
 int counter = 0;
 for (int z = x ^ y; z!= 0; z= z>>1) {
 counter += c & 1;
 }
 return counter;
 }

Reverse a Linked List

This question is one of the most repeated ones across Google, Microsoft, Amazon and a few more. The thing which makes it a good question is that after you scratch your head and give them an algorithm they either ask you to optimize it for the boundary cases (which is fine for most of the candidates) or they ask you to do it the other way round.

I am going to give you both, recursive as well as an iterative solution. For both the techniques :

Initially call the method as : head = reverseList(head,NULL)


Iterative :

Node *Reverse (Node *p)
{
   Node *pr = NULL;
   while (p != NULL)
   {
      Node *tmp = p->next;
      p->next = pr;
      pr = p;
      p = tmp;
   }
   return pr;
}
-----------------------------------------------------------------------
Recursive:

Node *reverseList(Node *current, Node *parent)
{
        Node *revHead;
   if(current == null)
      revHead = parent;
   else
   {
      revHead = reverseList(current->next,current);
      current->next = parent;      
   }
   return revHead;
}

Acyclic Linked-List Detection Algorithm

Write a function that takes a pointer to the head of a list and determines whether the list is cyclic or acyclic. Your function should return false if the list is acyclic and true if it is cyclic. You may not modify the list in any way.

int determineTermination( node *head )
        { 
           node *fast, *slow;
           fast = slow = head; 
                  while( true )
                       {
                           if( !fast || !fast->next ) 
                              return false; 
                           else if( fast == slow || fast->next == slow ) 
                              return true; 
                           else 
                              {
                                 slow = slow->next; 
                                 fast = fast->next->next; 
                              }
                        }
        }


Leave your comments below the post.
Happy coding :)

One line of code for finding out whether a number is power of 2 or not

(An obnoxious looking piece of code) 

If the following LOC returns true, find out what is x here ?

if((x&&!(x&(x-1))==1)) 

Study the Bit-wise operators well, many problems and contests have questions which become quite easy if these are used !!
 The above code is a simple one line test for testing if a variable entered or passed is a power of 2.

Amazon easy simple output question

Another of internet giant Amazon's technical written test question. This question tries to make filter out candidates who have come to attend the examination after reading refreshers of C language. Delve into your K&R's to figure it out.

main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}

Hints:
\n - newline
\b - backspace
\r - linefeed
.
.
.
.
Solution coming soon !!