3. Linear Structures

A computer processes some data. Always. Now what we are interested in this chapter is the structure of the data. Typically we will study linear structures like arrays, linked lists, stacks, queues etc. The structure of data is important because it governs how we can perform insertion, deletion, update and more such operations on data. It not only governs but also how fast we will be able to do such stuff. Some languages which have big standard libraries like C++, Python, Java etc they provide many utilities data structures and most common operations as a part of it. However, since C99 is having a much smaller standard library such facilities are not there.

It is assumed that the reader i.e. you know about basics of concepts like arrays, pointers, functions and recursion. C is primarily used to implement the data structures and algorithms therefore if you do not know C then probably you would want to read this.

Some of the linear structures are linked lists, stacks and queues. We will start with linked lists. Linked lists are also simply known as lists. A list can be mathematically modeled as \(t_1, t_2, ..., t_n\) i.e. a list having n items of element type. \(t_i\) is known as node at position \(i\) where \(i\) is used to indicate the position. All the elements are of the same type.

3.1. Singly Linked List

Let us begin with singlly linked lists. When we say linked typically we mean singly linked list. There is other types as well. Doubly linked lists and circular linked list for example. For now we will start with singly linked list because it is simplest to understand and can be used for as a base data structure to implement stacks and queues which are going to be coming chapters. A singly linked list looks something like given below:

\node at (0, 0) [rectangle, draw] (A1) {10};
\node at (2, 0) [rectangle, draw] (A2) {20};
\node at (4, 0) [rectangle, draw] (A3) {30};
\node at (6, 0) [rectangle, draw] (A4) {40};

\draw[<-, >=stealth] (A1.west) -- ++(-.5, 0);
\draw[->, >=stealth] (A1.east) -- (A2.west);
\draw[->, >=stealth] (A2.east) -- (A3.west);
\draw[->, >=stealth] (A3.east) -- (A4.west);
\draw[->, >=stealth] (A4.east) -- ++(.7, 0);

\draw (A1.west)+(-1,0) node {head};
\draw (A2.west)+(-.7,.2) node {next};
\draw (A3.west)+(-.7,.2) node {next};
\draw (A4.west)+(-.7,.2) node {next};
\draw (A4.east)+(.5,.2) node {next};
\draw (A4.east)+(1.5,0) node {NULL};

An example linked list.

A linked list’s first element is typically known as head and last element is known as tail. All nodes have a pointer which points to next node so that you can traverse list in forward direction. Note that it is impossible to traverse in opposite direction. tail’s next pointer point to NULL which indicates the ending point of the linked list. tail is not mandatory to have when dealing with linked list but a pointer to head is aleways present.

Impementation of a linked list involved a self referencing structure. Given below is a typical self referencing structure:

struct S {
  struct S *next;
};

struct S* head;

The presence of next which is a pointer of C causes some meory to be wasted in linear order equal to number of nodes in linked list. Note that real implementations will have some data members as well. Some common operation on a linked list are insertion at beginning, in between somewhere and at the end. We also can have an operation for counting number of nodes. Deletion and searching of a node is also there. Note that to do any of these operations except insertion at beginning we need to search the appropriate node which is an operation involving traversal of list in linear fashion. Therefore time compexity of all such operations is \(O(n)\). Insertion at beginning will have \(O(1)\) time complexity. However, if search operation for a particular node is to be considered separate then deletion and insertion at any place will have time complexity of \(O(n)\).

Let us implement a linked list and its operations:

#include <stdio.h>
#include <stdlib.h>

typedef struct linked_list {
    int data;
    struct linked_list *next;
}ll;

void menu();
void add_at_beg(ll**);
void print(ll*);
void search(ll*);
void delete(ll**);
void append(ll**);
void add_in_bet(**);

void delete(ll** head)
{
    ll *temp, *q;
    int i;

    temp = *head;

    if(*head == NULL) {
        printf("There is no element to be deleted.\n");
        return;
    }

    printf("Enter the value of data to be deleted.\n");
    scanf("%d", &i);

    if(temp->data == i) {
        *head = temp->next;
        free(temp);
        return;
    } else {
        while(temp->next != NULL) {
            if(temp->data == i) {
                q->next = temp->next;
                free(temp);
                return;
            }
            q = temp;
            temp = temp->next;
        }
    }

    if(temp->data == i) { // case of deletion of last node
        q->next = temp->next;
        free(temp);
        return;
    }

    printf("The element to be deleted was not found.\n");
}

int count(ll* head)
{
    int count = 1;

    if(head == NULL) {
        printf("The no. of elements in linked list is %d.\n", 0);
        return 0;
    }

    while(head->next != NULL) {
        count++;
        head = head->next;
    }

    printf("The no. of elements in linked list is %d.\n", count);
    return count;
}

void menu()
{
    puts("1. Add an element at beginning.");
    puts("2. Add an element at position n.");
    puts("3. Add an element at end.");
    puts("4. Count the number of elements.");
    puts("5. Delete an element.");
    puts("6. Search an element.");
    puts("7. Print the list.");
}

void append(ll** head)
{
    ll* temp, *q = *head;
    int i;

    printf("Enter the number which is to be appended to the list.\n");
    scanf("%d", &i);

    temp = (ll*)malloc(sizeof(ll));
    temp->data = i;
    temp->next = NULL;

    if(!(*head)) {
        *head=temp;
        return;
    }

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

    q->next = temp;
}

void add_in_bet(ll** head)
{
    ll *temp, *q = *head;
    int i = 0, j = 0;
    int position = 0;

    printf("Enter position at which the number is to be added.\n");
    scanf("%d", &position);

    if(position == 0)
        return add_at_beg(head);

    temp = (ll*)malloc(sizeof(ll));

    printf("Enter an integer to be added in between.\n");
    scanf("%d", &i);

    while(q->next != NULL) {
        ++j;
        if(j == position) {
            temp->next = q->next;
            q->next = temp;
            temp->data = i;
            return;
        }
        q = q->next;
    }
    ++j;
    // This is the case when q->next is NULL so it is an append
    // operation
    if(j == position) {
        append(head);
    }
    free(temp); // no insertion happened so we need to free temp
                // i.e. j was out of possible positions
}

void add_at_beg(ll** head)
{
    ll *temp;
    int i;

    temp = (ll*)malloc(sizeof(ll));

    printf("Enter an integer to be added at beginning\n");
    scanf("%d", &i);

    temp->next = *head;
    *head = temp;
    (*head)->data = i;
}

void print(ll* head)
{
    printf("Head-->");
    while(head != NULL) {
        printf("%d--->", head->data);
        head = head->next;
    }

    printf("NULL\n");
}

void search(ll* head)
{
    int i=0, position=1;

    printf("Enter the number to be searched.");
    scanf("%d", &i);

    while(head != NULL) {
        if(head->data == i) {
            printf("%d is found at %dth position.\n", i, position - 1);
            return;
        }

        head = head->next;
        position++;
    }
    printf("%d was not found in linked list.\n", i);
}

int main()
{
    ll* head = NULL;
    int option = 0;

    menu();
    printf("Enter 1 to 7 to choose an action. Any other number to quit.\n");
    scanf("%d", &option);
    getchar(); // to remove \n

    while(option  >= 1 && option <= 7) {
        switch(option) {
            case 1:
                add_at_beg(&head);
                break;
            case 2:
                add_in_bet(&head);
                break;
            case 3:
                append(&head);
                break;
            case 4:
                count(head);
                break;
            case 5:
                delete(&head);
                break;
            case 6:
                search(head);
                break;
            case 7:
                print(head);
                break;
            default:
                break;
        }
        menu();
        printf("Enter 1 to 7 to choose an action. Any other number to quit.\n");
        fflush(stdin);
        scanf("%d", &option);
        getchar(); // to remove \n
    }

    return 0;
}

Now I will explain these function one by one using images. First we discuss add_at_begin. Note that we can wrap all insertion functions by calling single insert function of the type insert(ll* head, int item, size_t position). Please note that I have used size_t for position because I want the list to be able to have as many members as malloc allows. If we use something like int which is nothing but signed int then we would be restricted to 2 * 1024 * 1024 * 1024 or 2147483648 members. Note that size_t is nothing but unsigned long which is 4 bytes on 32-bit systems and 8 bytes on 64-bit systems.

3.1.1. Insertion at the Beginning

Insertion at beginning is simple. We create a new node. Then we make its next pointer to point to current head and then use current head pointer to point to this new node. The entire operation is shown graphically below:

\node at(0, 0) [rectangle, draw] (A) {G};
\draw[->, >=stealth] (A.east) -- ++(1, 0);
\draw (A.east)+(.5, .2) node (B) {next};
\draw (A.north)+(0, 1) node (C) {temp};
\draw[->, >=stealth] (C.south) -- (A.north);
\draw (A.west)+(-4,0) node(D) {*head};
\draw[->, >=stealth] (D.east) -- ++(1, 0);
\draw (D.east)+(1.8, 0) node (E) {$NULL$};

\node [label={[align=center, yshift=-2.5cm]Initially $*head$ is $NULL$. We allocate a
$temp$ node.\\$temp->data$ contains garbage and $temp->next$ points to
unknown location.\\Let us say we want to insert 10.}] (F) {};

\node at(0, 0) [rectangle, draw] (A) {G};
\draw[->, >=stealth] (A.east) -- ++(1, 0);
\draw (A.east)+(.5, .2) node (B) {next};
\draw (A.north)+(0, 1) node (C) {temp};
\draw[->, >=stealth] (C.south) -- (A.north);
\draw (A.west)+(-4,0) node(D) {*head};
\draw[->, >=stealth] (D.east) -- ++(1, 0);
\draw (D.east)+(1.8, 0) node (E) {$NULL$};
\draw (A.east)+(1.8, 0) node {$NULL$};

\node [label={[align=center, yshift=-1.5cm]$temp->next$ is assigned $*head$
pointer which is $NULL$.}] (F) {};

\node at(0, 0) [rectangle, draw] (A) {10};
\draw[->, >=stealth] (A.east) -- ++(1, 0);
\draw (A.east)+(.5, .2) node (B) {next};
\draw (A.north)+(0, 1) node (C) {temp};
\draw[->, >=stealth] (C.south) -- (A.north);
\draw[<-, >=stealth] (A.west) -- ++(-1, 0);
\draw (A.west)+(-1.6, 0) node (E) {*head};
\draw (A.east)+(1.8, 0) node {$NULL$};

\node [label={[align=center, yshift=-1.5cm]$*head$ is assigned $temp$ and 10
is copied.}] (F) {};

\node at(-2, 0) [rectangle, draw] (A) {G};
\draw[->, >=stealth] (A.east) -- ++(1, 0);
\draw (A.east)+(.5, .2) node (B) {next};
\draw (A.north)+(0, 1) node (C) {temp};
\draw[->, >=stealth] (C.south) -- (A.north);

\node at(3, 0) [rectangle, draw] (D) {10};
\draw[<-, >=stealth] (D.west) -- ++(-1,0);
\draw (D.west)+(-1.6, 0) node (E) {*head};
\draw (D.east)+(1.8, 0) node {$NULL$};
\draw[->, >=stealth] (D.east) -- ++(1,0);

\node [label={[align=center, yshift=-1.5cm] To insert another node 20 before
10 we allocate $temp$.}] (F) {};

\node at(-2, 0) [rectangle, draw] (A) {G};
\draw (A.east)+(.5, .2) node (B) {next};
\draw (A.north)+(0, 1) node (C) {temp};
\draw[->, >=stealth] (C.south) -- (A.north);

\node at ($(A.east)+(1.5, 0)$) [rectangle, draw] (D) {10};
\draw[->, >=stealth] (A.east) -- (D.west);
\draw (D.north)+(0, 1) node (E) {*head};
\draw[->, >=stealth] (E.south) -- (D.north);

\draw (D.east)+(1.8, 0) node {$NULL$};
\draw[->, >=stealth] (D.east) -- ++(1,0);

\node [label={[align=center, yshift=-1.5cm]$temp->next$ is assigned
$*head.$}] (F) {};

\node at(-2, 0) [rectangle, draw] (A) {20};
\draw (A.east)+(.5, .2) node (B) {next};
\draw (A.north)+(0, 1) node (C) {temp};
\draw[->, >=stealth] (C.south) -- (A.north);

\node at ($(A.east)+(1.5, 0)$) [rectangle, draw] (D) {10};
\draw[->, >=stealth] (A.east) -- (D.west);
\node at ($(A.west)+(-1.5, 0)$) (E) {*head};
\draw[->, >=stealth] (E.east) -- (A.west);

\draw (D.east)+(1.8, 0) node {$NULL$};
\draw[->, >=stealth] (D.east) -- ++(1,0);

\node [label={[align=center, yshift=-1.5cm]$*head$ is assigned $temp$ and
data 20 is copied.}] (F) {};

Insertion at beginning of a singly linked list.

3.1.2. Insertion at Some Position

If position is 0 then call insertion at beginning. Note that in the code it is assumed that position will be less than size of list. Because if position is more then temp is allocated but it is not linked in to the structure and will cause memory leak. We iterate in the while loop to reach the position and then we do similar pointer manipulation like in last case to insert a node in between which is shown using images below. Consider the linked list having three elements 10, 20, 30 and head poiting to first element. Let us say we want to insert another element 15 in between 10 and 20. We initialize q with head. If position is 0 then it is insertion at beginning and we call that function. Else we allocate memory for temp to hold 20 and follow next pointer on q and incrememnt loop index j till it is equal to position. So when position and j are one q is pointing at 10. At this moment temp is not connected to list and its next pointer is pointing to some arbitrary location. Then we connect it to q->next which is 20. Now all is left is to move q->next and point it to temp and then copy the data and break the loop.

\node at (0, 0) [rectangle, draw] (A) {G};
\draw [->, >=stealth](A.east) -- ++(1, 0);
\node at ($(A.east) + (.5, .2)$) (B) {next};
\draw [<-, >=stealth](A.west) -- ++(-1, 0);
\node at ($(A.west) - (1.5, 0)$) (C) {temp};

\node at (0, -1) [rectangle, draw] (D) {20};
\node at (2, -1) [rectangle, draw] (E) {30};
\node at (-2, -1) [rectangle, draw] (F) {10};

\draw [->, >=stealth] (D.east) -- (E.west);
\draw [->, >=stealth] (F.east) -- (D.west);
\draw [->, >=stealth] (E.east) -- ++(1, 0);
\draw [<-, >=stealth](F.south west) -- ++(-1, 0);
\draw [<-, >=stealth](F.north west) -- ++(-1, 0);

\node at ($(D.east) + (.5, .2)$) (G) {next};
\node at ($(E.east) + (.5, .2)$) (H) {next};
\node at ($(F.east) + (.5, .2)$) (I) {next};

\node at ($(F.north west) - (1.5, 0)$) (J) {q};
\node at ($(F.south west) - (1.5, 0)$) (k) {head};

\node [label={[align=center, yshift=-3.5cm]We have 3 nodes. Let us say we
want to insert 15 after 10.\\We allocate $temp$ which contains garbage and
points somewhere.\\$q$ is initialized with $*head$.}] (F) {};

\node at (0, 0) [rectangle, draw] (A) {G};
\draw [<-, >=stealth](A.west) -- ++(-1, 0);
\node at ($(A.west) - (1.5, 0)$) (C) {temp};

\node at (0, -1) [rectangle, draw] (D) {20};
\node at (2, -1) [rectangle, draw] (E) {30};
\node at (-2, -1) [rectangle, draw] (F) {10};

\draw [->, >=stealth] (D.east) -- (E.west);
\draw [->, >=stealth] (F.east) -- (D.west);
\draw [->, >=stealth] (E.east) -- ++(1, 0);
\draw [<-, >=stealth](F.south west) -- ++(-1, 0);
\draw [<-, >=stealth](F.north west) -- ++(-1, 0);

\draw [->, >=stealth](A.south) -- (D.north);
\node at ($(A.south)!.5!(D.north) + (.5, 0)$) (B) {next};

\node at ($(D.east) + (.5, .2)$) (G) {next};
\node at ($(E.east) + (.5, .2)$) (H) {next};
\node at ($(F.east) + (.5, .2)$) (I) {next};

\node at ($(F.north west) - (1.5, 0)$) (J) {q};
\node at ($(F.south west) - (1.5, 0)$) (k) {head};

\node [label={[align=center, yshift=-3cm]$q->next$ is not $NULL$. $j$
increases to 1. Position is 1.\\$temp->next$ is assigned $q->next$.}] (F)
{};

\node at (0, 0) [rectangle, draw] (A) {15};
\draw [<-, >=stealth](A.west) -- ++(-1, 0);
\node at ($(A.west) - (1.5, 0)$) (C) {temp};

\node at (0, -1) [rectangle, draw] (D) {20};
\node at (2, -1) [rectangle, draw] (E) {30};
\node at (-2, -1) [rectangle, draw] (F) {10};

\draw [->, >=stealth] (D.east) -- (E.west);
\draw [->, >=stealth] (F.north east) -- (A.south west);
\draw [->, >=stealth] (E.east) -- ++(1, 0);
\draw [<-, >=stealth](F.south west) -- ++(-1, 0);
\draw [<-, >=stealth](F.north west) -- ++(-1, 0);

\draw [->, >=stealth](A.south) -- (D.north);
\node at ($(A.south)!.5!(D.north) + (.5, 0)$) (B) {next};

\node at ($(D.east) + (.5, .2)$) (G) {next};
\node at ($(E.east) + (.5, .2)$) (H) {next};
\node at ($(F.east) + (.5, .2)$) (I) {next};

\node at ($(F.north west) - (1.5, 0)$) (J) {q};
\node at ($(F.south west) - (1.5, 0)$) (k) {head};

\node [label={[align=center, yshift=-3cm]$q->next$ is assigned $temp$ and
data is copied.\\And we break out of loop.}] (F)
{};

Insertion in between singly linked list.

3.1.3. Insertion at the End or Append

If the list is not empty append is easy just moving a few pointers at the end. Making the next of last node point to the newly allocated memory and next of newly allocated memory at NULL. However, if the list is empty is simply insetion at beginning.

3.1.4. Searching an Element

Perhaps one of the important operations because it also affects deletion or insetion before/after a particular node. It is ealiy implemented by starting from head examining data and proceeding to next untill a match is found.

3.1.5. Deleting an Element

This is similar to insertion just reverse in nature. If the node to be deleted is head node the we simply copy the name of head to temp and make head point to next element. Then we free temp. It is slightly more involving to delete if the node is not head. In that case we point current node’s next to temp's next and then delete temp. It is shown below using diagrams.

3.1.6. Counting the Size

Most list implementations provide a size attribute which is always updated upon insertion and deletion and can be used to know the current size of list. The count function is similar in functionality just that to maintain a size we will need to call it whenever a value is inserted or deleted. The implementation of count is easily doby by starting with head and incrementing a variable as long as end (marked by NULL) is not encountered.

3.2. Singly Linked List vs Array

For size based comparison an array is fixed while a singly linked list can grow as needed so there is no competition there. Just that allocating one node at a time may make things less efficient. However, this is also good in one sense that on a system low with RAM the problems caused by memory fragmentation will be less problematic.

For time complexity let us analyze that:

Operation Linked Lists Arrays
Search \(O(n)\) \(\Theta(1)\)
Insert/Delete at beginning \(\Theta(1)\) N/A
Insert/Delete at end \(\Theta(n)\) Last element is unknown \(\Theta(1)\) Last element is unknown N/A
Insert/Delete in middle search time + \(\Theta(1)\) N/A
Wasted space \(\Theta(n)\) 0

As we can see lists are on the slower side but deleting is problematic in arrays. In fixed arrays you can mark as deleted while in dynamic arrays you need to shift entire content which is not efficient. Lists also waste small amount of memory for pointer members which are needed to enforce the structure. However, the flexibility of linked lists is a big win. This is not to say that for all situations any of these are a sure-fire solution. It depends on the problem whether arrays should be used or lists should be used.

3.3. Questions on Singly Linked Lists

  1. Implement a stack using singly linked list(These is done in the chapter of Stacks),

  2. Implement a queue using singly linked list(This is done in the chapter of Queues).

  3. Reverse a singly linked list iteratively and recursively.

  4. Sort a linked list using bubble sort. (Look for sorting chapter or on internet for details on bubble sort)

  5. Sort a linked list using quick sort.(Look for sorting chapter or on internet for details on quick sort).

  6. Sort a linked list using merge sort.(Look for sorting chapter or on internet for details on merge sort).

  7. Detect a loop in a singly linked list. (HINT: This can happen only at last node whose next pointer will end up pointing to one of previous nodes.)

  8. Use singly linked lists to implement arbitrary-precision arithmetic library having basic functionality of addition, subtraction, multiplication and division. Fist do it for integers then do it for floats.

  9. Given two lists find their union and intersection.

  10. You are given a pointer to a node in a linked list and you have to delete it. Write a function to do that.

  11. Write a function to find the middle node of a given linked list.

  12. Write a function to find the nth node from the end.

  13. Write a function to check if a given list is a pallindrome.

  14. Write a function to insert in a list in a sorted way. The list is already sorted.

  15. There are two lists. One of the lists is linked to another making some part common for both. Write a function to find the intersection point.

  16. Write a function to remove duplicates from a sorted linked list.

  17. Write a function to remove duplicated from an unsorted linked list.

  18. Write a function to swap nodes of a list in pairs.

  19. Write a function to reverse every n nodes of a linked list.

  20. Write a function to delete all those nodes for which a greater value exist away from head after the node.

  21. Given three linked lists of integers write a function to find a triplets whose sum is equal to a given number.

  22. Write a function to rotate a linked list in anti-clockwise direction by n nodes.

  23. There is a linked list of linked lists. Every node contains two pointers. (i) Pointer to next node in the main list. (ii) Pointer to a linked list where this node is head.

    Write a function to convert it to a single list given all lists are sorted.

  24. Write a function to sort a linked list which contains only 0s, 1s and 2s.

  25. Given a linked list of co-ordinates where adjacent points either form a vertical line or a horizontal line. Delete points from the linked list which are in the middle of a horizontal or vertical line.

  26. Given a linked list where in addition to the next pointer, each node has a child pointer, which may or may not point to a separate list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure. You are given the head of the first level of the list. Flatten the list so that all the nodes appear in a single-level linked list. You need to flatten the list in way that all nodes at first level should come first, then nodes of second level, and so on.

  27. Given two sorted linked lists, construct a linked list that contains maximum sum path from start to end. The result list may contain nodes from both input lists. When constructing the result list, we may switch to the other input list only at the point of intersection (which mean the two node with the same value in the lists). You are allowed to use \(O(1)\) extra space.

  28. Farey fractions of level one are deined as sequence \((0/1, 1/1)\). For level 2 it is \((0/1, 1/2, 1/1)\), then for level 3 it is \((0/1, 1/3, 1/2, 2/3, 1/1)\). So at each level a new fraction \((a+b)/(c+d)\) is inserted beween the fractions \(a/c\) and \(b/d\) if \(c+d \le n\). Write a program which creates a linked list for level n by constantly extending it.

3.4. Doubly Linked Lists

Singly linked lists are pretty good but quite inflexible as far as traversal is concerned. You can traverse only in ofrward direction. We can improve this by having another pointer to each node prev which will point to previous node. Having this new prev pointer will allow us to traverse in both the directions which is very nice to have functionality in certain situations. The nature of doubly linked list provides another feature that is we can make useful circular lists.

You might have noticed that I have kept only a head pointer in the singly linked list representation and no tail pointer. The reason for having a tail pointer will help in appending at the end for sure. However, since we can move only in forward direction the tail pointer is useless in deletion. For deletion we will have to traverser from head in any case. But when we are designing a doubly linked list a tail pointer helps both in appending as well as deleting a node. We will see more of doubly linked lists in their respective chapter.