Swapping the odd and even position in a linked list

A simple and intuitive solution to the problem !! It might not be very efficient but it does certainly work.


node* swap(node *head)
{
    node dummy;
    node *prev, *curr, *u, *tmp;

    if (head==NULL) return head;

    prev=&dummy;
    for(curr=head; curr!=NULL && curr->next!=NULL; ) {
        u = curr->next;
        tmp = u->next;
        prev->next = u;
        u->next = curr;
        prev = curr;
        curr = tmp;
    }
    prev->next = curr;
    return dummy.next;
}

void print(const node *head)
{
    for(const node *p=head; p; p=p->next)
        printf("%d ", p->val);
    putchar('\n');
}

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;
}

Easy-4

void main()

{

int *mptr, *cptr;

mptr = (int*)malloc(sizeof(int));                               //Output: some garbage values

printf(“%d”,*mptr);

int *cptr = (int*)calloc(sizeof(int),1);                      // Output=0

printf(“%d”,*cptr);

}

The above statements return garbage values and a zero in the next.

So much for understanding a small difference between malloc() and calloc().

See the differences in arguments and the returned values.

A few simple logical questions !

1.) 

Multiplying a number by 7 without using * and + operators.

NewNum = Num << 3;                      // multiplied by 2^3 = 8

NewNum = NewNum - Num;             // 8 – 1 = 7

2.)

Finding the last character of any String.

                              (OR)

Write a function that finds the last instance of a character in a string


char *lastChar(char *String, char ch)
{
char *cStr = NULL;


// traverse the entire string


while(*String ++ != NULL ) 
{
if(*String==ch) 
cStr = String;
}
return cStr;
} 

C Output Type Question

main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
}

Try to find the output !!
This question is seemingly quite unconventional, but try it out without using a compiler, and then compare your results with:


1 1 1
2 2 2
3 3 3
3 4 4

Brain-Teaser

Try this without googling the answer.. also please try to justify your assumptions if any ?

A party of four travelers comes to a rickety bridge at night. The bridge can hold the weight of at most two of the travelers at a time, and it cannot be crossed without using a flashlight. The travelers have one flashlight among them. Each traveler walks at a different speed: The first can cross the bridge in 1 minute, the second in 2 minutes, the third in 5 minutes, and the fourth takes 10 minutes to cross the bridge. If two travelers cross together, they walk at the speed of the slower traveler.

What is the least amount of time in which all the travelers can cross from one side of the bridge to the other?

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 :)

Microsoft Moderate Programming Question

An important interview as well as optimised solution for finding the largest sub array sum in an array which may contain 0, positive and negative integers.


#include<stdio.h>
main()
{
int sumtillnow=0,sum=0,i=0;
int a[7]={-1,11,13,12,-90,23,5};
for(i=0;i<7;i++)
{
sumtillnow+=a[i];
if(sumtillnow>sum)
 sum=sumtillnow;
if(sumtillnow<0)
 sumtillnow=0;
 }
printf("%d",sum);
}

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 !!