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;
}

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);
}