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Question 1


How do you reverse a singly linked list? How do you reverse a doubly linked list? Write a C program to do the same.


This is THE most frequently asked interview question. The most!.

Singly linked lists

Here are a few C programs to reverse a singly linked list.



Method1 (Iterative)


#include 


// Variables
typedef struct node 

   int value; 
   struct node *next; 
}mynode;


// Globals (not required, though).
mynode *head, *tail, *temp; 


// Functions
void add(int value);
void iterative_reverse();
void print_list();


// The main() function
int main()
{
    head=(mynode *)0;
  
    // Construct the linked list.
    add(1);
    add(2);
    add(3);
  
    //Print it
    print_list();
  
    // Reverse it.
    iterative_reverse();

    //Print it again
    print_list();
  
    return(0);
}


// The reverse function
void iterative_reverse()
{
     mynode *p, *q, *r;
  
     if(head == (mynode *)0)
     {
        return;
     }

  
     p       = head;
     q       = p->next;
     p->next = (mynode *)0;
  

     while (q != (mynode *)0)
     {
        r       = q->next;
        q->next = p;
        p       = q;
        q       = r;
     }
   
     head = p;
}


// Function to add new nodes to the linked list
void add(int value)
{
   temp = (mynode *) malloc(sizeof(struct node));
   temp->next=(mynode *)0;
   temp->value=value;
   
   if(head==(mynode *)0)
   {
      head=temp;
      tail=temp;
   }
   else
   {
     tail->next=temp;
     tail=temp;
   }
}


// Function to print the linked list.
void print_list()
{
  printf("\n\n");
  for(temp=head; temp!=(mynode *)0; temp=temp->next)
  {
     printf("[%d]->",(temp->value));
  }
  printf("[NULL]\n\n");
}









Method2 (Recursive, without using any temporary variable)


#include 


// Variables
typedef struct node 

   int value; 
   struct node *next; 
}mynode;

// Globals.
mynode *head, *tail, *temp;


// Functions
void add(int value);
mynode* reverse_recurse(mynode *root);
void print_list();


// The main() function
int main()
{
    head=(mynode *)0;
  
    // Construct the linked list.
    add(1);
    add(2);
    add(3);
  
    //Print it
    print_list();

    // Reverse it.
    if(head != (mynode *)0)
    {
      temp = reverse_recurse(head);
      temp->next = (mynode *)0;
    }
  
    //Print it again
    print_list();
  
    return(0);
}


// Reverse the linked list recursively
//

// This function uses the power of the stack to make this
// *magical* assignment
//
// node->next->next=node; 
//
// :)

mynode* reverse_recurse(mynode *root)

  if(root->next!=(mynode *)0) 
  { 
      reverse_recurse(root->next); 
      root->next->next=root; 
      return(root); 
  } 
  else 
  { 
      head=root; 
  } 




// Function to add new nodes to the linked list.
void add(int value)
{
     temp = (mynode *) malloc(sizeof(struct node));
     temp->next=(mynode *)0;
     temp->value=value;
   
     if(head==(mynode *)0)
     {
        head=temp;
        tail=temp;
     }
     else
     {
       tail->next=temp;
       tail=temp;
     }
}


// Function to print the linked list.
void print_list()
{
    printf("\n\n");
    for(temp=head; temp!=(mynode *)0; temp=temp->next)
    {
       printf("[%d]->",(temp->value));
    }
    printf("[NULL]\n\n");
}










Method3 (Recursive, but without ANY global variables. Slightly messy!)


#include 


// Variables
typedef struct node 

   int value; 
   struct node *next; 
}mynode;


// Functions
void add(mynode **head, mynode **tail, int value);
mynode* reverse_recurse(mynode *current, mynode *next);
void print_list(mynode *);


int main()
{
    mynode *head, *tail;
    head=(mynode *)0;
  
    // Construct the linked list.
    add(&head, &tail, 1);
    add(&head, &tail, 2);
    add(&head, &tail, 3);
  
    //Print it
    print_list(head);

    // Reverse it.
    head = reverse_recurse(head, (mynode *)0);
  
    //Print it again
    print_list(head);
  
    getch();
    return(0);
}


// Reverse the linked list recursively
mynode* reverse_recurse(mynode *current, mynode *next)

  mynode *ret;
  
  if(current==(mynode *)0)
  {
    return((mynode *)0);
  }
  
  ret = (mynode *)0;
  if (current->next != (mynode *)0)
  {
     ret = reverse_recurse(current->next, current);
  }
  else
  {
     ret = current;
  }
 
  current->next = next;
  return ret;



// Function to add new nodes to the linked list.
// Takes pointers to pointers to maintain the 
// *actual* head and tail pointers (which are local to main()).

void add(mynode **head, mynode **tail, int value)
{
     mynode *temp1, *temp2;
     
     temp1 = (mynode *) malloc(sizeof(struct node));
     temp1->next=(mynode *)0;
     temp1->value=value;
   
     if(*head==(mynode *)0)
     {
        *head=temp1;
        *tail=temp1;
     }
     else
     {
        for(temp2 = *head; temp2->next!= (mynode *)0; temp2=temp2->next);
        temp2->next = temp1;
        *tail=temp1;
     }
}


// Function to print the linked list.
void print_list(mynode *head)
{
    mynode *temp;
    printf("\n\n");
    for(temp=head; temp!=(mynode *)0; temp=temp->next)
    {
       printf("[%d]->",(temp->value));
    }
    printf("[NULL]\n\n");
}








Doubly linked lists


This is really easy, just keep swapping the prev and next pointers and at the end swap the head and the tail:)


#include

#include


typedef struct node
{
  int value;
  struct node *next;
  struct node *prev;
}mynode ;

mynode *head, *tail;
void add_node(int value);
void print_list();
void reverse();

int main()
{



 head=NULL;
 tail=NULL;

 add_node(1);
 add_node(2);
 add_node(3);
 add_node(4);
 add_node(5);

 print_list();

 reverse();

 print_list();

 return(1);

}


void add_node(int value)
{
  mynode *temp, *cur;
  temp = (mynode *)malloc(sizeof(mynode));
  temp->next=NULL;
  temp->prev=NULL;

  if(head == NULL)
  {

    printf("\nAdding a head pointer\n");
    head=temp;
    tail=temp;
    temp->value=value;

  }
  else
  {



   for(cur=head;cur->next!=NULL;cur=cur->next);
   cur->next=temp;
   temp->prev=cur;
   temp->value=value;
   tail=temp;

  }

}


void print_list()
{
  mynode *temp;

  printf("\n--------------------------------\n");
  for(temp=head;temp!=NULL;temp=temp->next)
  {
    printf("\n[%d]\n",temp->value);
  }

}



void reverse()
{
  mynode *cur, *temp, *save_next;
  if(head==tail)return;
  if(head==NULL || tail==NULL)return;
  for(cur=head;cur!=NULL;)
  {
    printf("\ncur->value : [%d]\n",cur->value);
    temp=cur->next;
    save_next=cur->next;
    cur->next=cur->prev;
    cur->prev=temp;
    cur=save_next;
  }

  temp=head;
  head=tail;
  tail=temp;
}





Having shown all these different methods, if someone can mail me a really, really good practical application of reversing a linked list (singly or doubly linked list), I would be really thankful to them. I have not found one good application of this. All I see is an urge to understand how well the candidate handles the pointer manipulation.


Question 2

Given only a pointer to a node to be deleted in a singly linked list, how do you delete it?


This is a very good interview question

The solution to this is to copy the data from the next node into this node and delete the next node!. Ofcourse this wont work if the node to be deleted is the last node. Mark it as dummy in that case. If you have a Circular linked list, then this might be all the more interesting. Try writing your own C program to solve this problem. Having a doubly linked list is always better.


Question 3

How to create a copy of a linked list? Write a C program to create a copy of a linked list.


Check out this C program which creates an exact copy of a linked list.


copy_linked_lists(struct node *q, struct node **s)
{
    if(q!=NULL)
    {
        *s=malloc(sizeof(struct node));
        (*s)->data=q->data;
        (*s)->link=NULL;
        copy_linked_list(q->link, &((*s)->link));
    }
}



Question 4

Write a C program to free the nodes of a linked list


Before looking at the answer, try writing a simple C program (with a for loop) to do this. Quite a few people get this wrong.


This is the wrong way to do it


struct list *listptr, *nextptr;
for(listptr = head; listptr != NULL; listptr = listptr->next) 
{
  free(listptr);
}



If you are thinking why the above piece of code is wrong, note that once you free the listptr node, you cannot do something like listptr = listptr->next!. Since listptr is already freed, using it to get listptr->next is illegal and can cause unpredictable results!



This is the right way to do it


struct list *listptr, *nextptr;
for(listptr = head; listptr != NULL; listptr = nextptr) 
{
  nextptr = listptr->next;
  free(listptr);
}



Question 5

Can we do a Binary search on a linked list?


Great C datastructure question!

The answer is ofcourse, you can write a C program to do this. But, the question is, do you really think it will be as efficient as a C program which does a binary search on an array?

Think hard, real hard.

Do you know what exactly makes the binary search on an array so fast and efficient? Its the ability to access any element in the array in constant time. This is what makes it so fast. You can get to the middle of the array just by saying array[middle]!. Now, can you do the same with a linked list? The answer is No. You will have to write your own, possibly inefficient algorithm to get the value of the middle node of a linked list. In a linked list, you loosse the ability to get the value of any node in a constant time.

One solution to the inefficiency of getting the middle of the linked list during a binary search is to have the first node contain one additional pointer that points to the node in the middle. Decide at the first node if you need to check the first or the second half of the linked list. Continue doing that with each half-list.


Question 6

Write a C program to return the nth node from the end of a linked list.


Here is a solution which is often called as the solution that uses frames.

Suppose one needs to get to the 6th node from the end in this LL. First, just keep on incrementing the first pointer (ptr1) till the number of increments cross n (which is 6 in this case)


STEP 1    :   1(ptr1,ptr2) -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10

STEP 2    :   1(ptr2) -> 2 -> 3 -> 4 -> 5 -> 6(ptr1) -> 7 -> 8 -> 9 -> 10




Now, start the second pointer (ptr2) and keep on incrementing it till the first pointer (ptr1) reaches the end of the LL.


STEP 3    :   1 -> 2 -> 3 -> 4(ptr2) -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 (ptr1)


So here you have!, the 6th node from the end pointed to by ptr2!


Here is some C code..


struct node
{
  int data;
  struct node *next;
}mynode;


mynode * nthNode(mynode *head, int n /*pass 0 for last node*/)
{
  mynode *ptr1,*ptr2;
  int count;

  if(!head)
  {
    return(NULL);
  }

  ptr1  = head;
  ptr2  = head;
  count = 0;

  while(count    {
     count++;
     if((ptr1=ptr1->next)==NULL)
     {
        //Length of the linked list less than n. Error.
        return(NULL);
     }
  }

  while((ptr1=ptr1->next)!=NULL)
  {
    ptr2=ptr2->next;
  }

  return(ptr2);
}



Question 7

How would you find out if one of the pointers in a linked list is corrupted or not?


This is a really good interview question. The reason is that linked lists are used in a wide variety of scenarios and being able to detect and correct pointer corruptions might be a very valuable tool. For example, data blocks associated with files in a file system are usually stored as linked lists. Each data block points to the next data block. A single corrupt pointer can cause the entire file to be lost!


 

  • Discover and fix bugs when they corrupt the linked list and not when effect becomes visible in some other part of the program. Perform frequent consistency checks (to see if the linked list is indeed holding the data that you inserted into it).
  • It is good programming practice to set the pointer value to NULL immediately after freeing the memory pointed at by the pointer. This will help in debugging, because it will tell you that the object was freed somewhere beforehand. Keep track of how many objects are pointing to a object using reference counts if required.
  • Use a good debugger to see how the datastructures are getting corrupted and trace down the problem. Debuggers like ddd on linux and memory profilers like Purify, Electric fence are good starting points. These tools should help you track down heap corruption issues easily.
  • Avoid global variables when traversing and manipulating linked lists. Imagine what would happen if a function which is only supposed to traverse a linked list using a global head pointer accidently sets the head pointer to NULL!.
  • Its a good idea to check the addNode() and the deleteNode() routines and test them for all types of scenarios. This should include tests for inserting/deleting nodes at the front/middle/end of the linked list, working with an empty linked list, running out of memory when using malloc() when allocating memory for new nodes, writing through NULL pointers, writing more data into the node fields then they can hold (resulting in corrupting the (probably adjacent) "prev" and "next" pointer fields), make sure bug fixes and enhancements to the linked list code are reviewed and well tested (a lot of bugs come from quick and dirty bug fixing), log and handle all possible errors (this will help you a lot while debugging), add multiple levels of logging so that you can dig through the logs. The list is endless...
  • Each node can have an extra field associated with it. This field indicates the number of nodes after this node in the linked list. This extra field needs to be kept up-to-date when we inserte or delete nodes in the linked list (It might become slightly complicated when insertion or deletion happens not at end, but anywhere in the linked list). Then, if for any node, p->field > 0 and p->next == NULL, it surely points to a pointer corruption.
  • You could also keep the count of the total number of nodes in a linked list and use it to check if the list is indeed having those many nodes or not.


The problem in detecting such pointer corruptions in C is that its only the programmer who knows that the pointer is corrupted. The program has no way of knowing that something is wrong. So the best way to fix these errors is check your logic and test your code to the maximum possible extent. I am not aware of ways in C to recover the lost nodes of a corrupted linked list.

I have a hunch that interviewers who ask this question are probably hinting at something called Smart Pointers in C++. Smart pointers are particularly useful in the face of exceptions as they ensure proper destruction of dynamically allocated objects. They can also be used to keep track of dynamically allocated objects shared by multiple owners. This topic is out of scope here, but you can find lots of material on the Internet for Smart Pointers.

If you have better answers to this question, let me know!


Question 8

How do you sort a linked list? Write a C program to sort a linked list.


This is a very popular interview question, which most people go wrong. The ideal solution to this problem is to keep the linked list sorted as you build it. This really saves a lot of time which would have been required to sort it.

However....

Method1 (Usual method)

The general idea is to decide upon a sorting algorithm (say bubble sort). Then, one needs to come up with different scenarios to swap two nodes in the linked list when they are not in the required order. The different scenarios would be something like


1. When the nodes being compared are not adjacent and one of them is the first node.
2. When the nodes being compared are not adjacent and none of them is the first node
3. When the nodes being compared are adjacent and one of them is the first node.
4. When the nodes being compared are adjacent and none of them is the first node.




One example bubble sort for a linked list goes like this


for(i = 1; i  {
    p1 = head;
    p2 = head->next;
    p3 = p2->next;
  
    for(j = 1; j <= (n - i); j++)
    {
       if(p2->value value)  
       {
           p2->next = p3->next;
           p3->next = p2;
           p1->next = p3;
           p1       = p3;
           p3       = p2->next;
       }
       else
       {
           p1 = p2;
           p2 = p3;
           p3 = p3->next;
       }
    }
}



As you can see, the code becomes quite messy because of the pointer logic. Thats why I have not elaborated too much on the code, nor on variations such as soring a doubly linked list. You have to do it yourself once to understand it.



Method1 (Divide and Conquer using merge sort)

The pseudocode for this method is



typedef struct node
{
   int value;
   struct node *next;
}mynode;

mynode *head, *tail;
int size;

mynode *mergesort(mynode *list, int size);
void display(mynode *list);

mynode *mergesort(mynode *list, int size)
{
  int size1, size2;
  mynode *tempnode1, *tempnode2, *tempnode3;

  if(size<=2)
  {
     if(size==1)
     { 
         // Nothing to sort!
         return(list);
     }
     else
     {
         if(list->value next->value
         {
            // These 2 nodes are already in right order, no need to sort
            return(list);
         }
         else
         {
            // Need to swap these 2 nodes
            /* Here we have 2 nodes
             * 
             * node 1 -> node2 -> NULL
             * 
             * This should be converted to
             * 
             * node2 -> node1 -> NULL
             *
             */

             tempnode1 = list;
             tempnode2 = list->next;
             tempnode2->next = tempnode1;
             tempnode1->next = NULL;
             return(tempnode2);
         }
     }   
  }
  else
  {
      // The size of the linked list is more than 2.
      // Need to split this linked list, sort the 
      // left and right sub-linked lists and merge.
  
      // Split.
      // tempnode1 will have the first half of the linked list of size "size1".
      // tempnode2 will have the second half of the linked list of size "size2".

      

      // Sort the two halves recursively 

      tempnode1 = mergesort(tempnode1, size1);
      tempnode2 = mergesort(tempnode2, size2);

      // Now merge the sorted lists back, let tempnode3 point to that new list.

      
      
      return(tempnode3);
  }
}





The code to merge the two already sorted sub-linked lists into a sorted linked list could be something like this..


mynode * merge(mynode *a, mynode *b)
{
   mynode *i, *j, *k, *c;

   i = a;
   j = b;
   c = getNewNode();
   k = getNewNode();

   while(i!=NULL && j!=NULL)
   {
     if(i->value value)
     {
        k->next=i;
        i=i->next;
     }
     else
     {
        k->next=j;
        j=j->next;
     }
   }
   
   if(i!=NULL)
       k->next=i;
   else
       k->next=j;

   return(c->next);

}



Question 9

How to declare a structure of a linked list?
 


The right way of declaring a structure for a linked list in a C program is


struct node {
  int value;
  struct node *next;
};
typedef struct node *mynode;



Note that the following are not correct


typedef struct {
  int value;
  mynode next;
} *mynode;



The typedef is not defined at the point where the "next" field is declared.



struct node {
  int value;
  struct node next;
};
typedef struct node mynode;



You can only have pointer to structures, not the structure itself as its recursive!


Question 10

Write a C program to implement a Generic Linked List.
 


Here is a C program which implements a generic linked list. This is also one of the very popular interview questions thrown around. The crux of the solution is to use the void C pointer to make it generic. Also notice how we use function pointers to pass the address of different functions to print the different generic data.


#include 

#include 

#include 


typedef struct list {
    void *data;
    struct list *next;
} List;

struct check {
    int i;
    char c;
    double d;
} chk[] = { { 1, 'a', 1.1 },
          { 2, 'b', 2.2 },
          { 3, 'c', 3.3 } };

void insert(List **, void *, unsigned int);
void print(List *, void (*)(void *));
void printstr(void *);
void printint(void *);
void printchar(void *);
void printcomp(void *);

List *list1, *list2, *list3, *list4;

int main(void)
{
    char c[]    = { 'a', 'b', 'c', 'd' };
    int i[]     = { 1, 2, 3, 4 };
    char *str[] = { "hello1", "hello2", "hello3", "hello4" };

    list1 = list2 = list3 = list4 = NULL;

    insert(&list1, &c[0], sizeof(char));
    insert(&list1, &c[1], sizeof(char));
    insert(&list1, &c[2], sizeof(char));
    insert(&list1, &c[3], sizeof(char));

    insert(&list2, &i[0], sizeof(int));
    insert(&list2, &i[1], sizeof(int));
    insert(&list2, &i[2], sizeof(int));
    insert(&list2, &i[3], sizeof(int));

    insert(&list3, str[0], strlen(str[0])+1);
    insert(&list3, str[1], strlen(str[0])+1);
    insert(&list3, str[2], strlen(str[0])+1);
    insert(&list3, str[3], strlen(str[0])+1);

    insert(&list4, &chk[0], sizeof chk[0]);
    insert(&list4, &chk[1], sizeof chk[1]);
    insert(&list4, &chk[2], sizeof chk[2]);

    printf("Printing characters:");
    print(list1, printchar);
    printf(" : done\n\n");

    printf("Printing integers:");
    print(list2, printint);
    printf(" : done\n\n");

    printf("Printing strings:");
    print(list3, printstr);
    printf(" : done\n\n");

    printf("Printing composite:");
    print(list4, printcomp);
    printf(" : done\n");

    return 0;
}

void insert(List **p, void *data, unsigned int n)
{
    List *temp;
    int i;

    /* Error check is ignored */
    temp = malloc(sizeof(List));
    temp->data = malloc(n);
    for (i = 0; i          *(char *)(temp->data + i) = *(char *)(data + i);
    temp->next = *p;
    *p = temp;
}

void print(List *p, void (*f)(void *))
{
    while (p) 
    {
        (*f)(p->data);
        p = p->next;
    }
}

void printstr(void *str)
{
    printf(" \"%s\"", (char *)str);
}

void printint(void *n)
{
    printf(" %d", *(int *)n);
}

void printchar(void *c)
{
    printf(" %c", *(char *)c);
}

void printcomp(void *comp)
{
    struct check temp = *(struct check *)comp;
    printf(" '%d:%c:%f", temp.i, temp.c, temp.d);
}








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