Skip to main content

Queue data structure

Queue is a linear data structure to which values are added and removed using FIFO method - first in first out. That is to say, the value which is added first, will be removed first from the queue. Exactly like the behavior of a real life queue.

The other related data structure is stack, from which elements are added and removed using LIFO method - last in first out.

Terminology





An element is always added to the rear end of the queue. This operation is called enqueue

An element is added from the front end of the queue. The operation of removing an element is called dequeue operation.


Implementation of queue

You can implement a queue using an array or a linked list. Array poses the problem that it is limited by its size. But by making the array as circular, queue can be implemented. Here when the array index reaches maximum value, queue insertion happens at the beginning of the array.

But implementation of a queue with linked list is not having such limitations. If you just add one more pointer for front end of the queue, all operations can be implemented with O(1) complexity.

Let us look at the algorithm for queue using linked list

  • Enqueue operation
    • create a new node
    • if rear is NULL, rear = front = newnode
    • else link rear node to newnode
    • set rear as newnode
  •  Dequeue operation
    • if queue is empty return error code
    • set temp to front
    • set front to next node of front
    • free the memory allocated to front node
    • return the value of temp
  • IsEmpty
    •  if front = null and rear = null return true
Do not forget the initialize the queue by setting front and rear equal to NULL. You don't want to get segmentation fault.

Enqueue operation


NODEPTR create_node(int n)
{
NODEPTR newnode = malloc(sizeof(struct node));
newnode->num = n;
newnode->next = NULL;
return newnode;
}

void enqueue(QUEUE *q1,int n)
{
NODEPTR temp = create_node(n);
if(is_empty(q1))
{
q1->rear = q1->front = temp;
}
else{
q1->rear->next = temp;
q1->rear = temp;
}
}

Next let us look at dequeue function


int is_empty(QUEUE *q1)
{
return (q1->front==NULL && q1->rear==NULL);
}
int dequeue(QUEUE *q1)
{
if(is_empty(q1))
return -1;

NODEPTR temp = q1->front;
int m = temp->num;
q1->front = q1->front->next;
free(temp);
if(q1->front==NULL)
q1->rear = NULL;
return m;
}

I have added an additional line in dequeue function, if the q1->front = NULL then even q1->rear = NULL, because when all the nodes are removed from the queue, both pointers must be NULL.

And here is the complete program.

#include<stdio.h>
#include<stdlib.h>
struct node
{
int num;
struct node *next;
};

typedef struct node *NODEPTR;

struct queue
{
NODEPTR front,rear;
};

typedef struct queue QUEUE;


NODEPTR create_node(int n)
{
NODEPTR newnode = malloc(sizeof(struct node));
newnode->num = n;
newnode->next = NULL;
return newnode;
}
int is_empty(QUEUE *q1)
{
return (q1->front==NULL && q1->rear==NULL);
}
void enqueue(QUEUE *q1,int n)
{
NODEPTR temp = create_node(n);
if(is_empty(q1))
{
q1->rear = q1->front = temp;
}
else{
q1->rear->next = temp;
q1->rear = temp;
}
}

int dequeue(QUEUE *q1)
{
if(is_empty(q1))
return -1;

NODEPTR temp = q1->front;
int m = temp->num;
q1->front = q1->front->next;
free(temp);
if(q1->front==NULL)
q1->rear = NULL;
return m;
}

void display(QUEUE q1)
{
NODEPTR node=q1.front;
printf("\nThe list is \n");
while(node !=NULL)
{
printf("%d----",node->num);
node = node->next;
}
}

int main()
{
QUEUE q1;
/*initialize queue*/
q1.front = q1.rear = NULL;

while(1)
{
int ans;
printf("Enter 1 - Add a node 2 - remove a node 3 - quit\n");
scanf("%d",&ans);
if (ans==1)
{
int n;
printf("Node to enqueued:");
scanf("%d",&n);
enqueue(&q1,n);
display(q1);
}else if (ans==2)
{
int n;
n = dequeue(&q1);
if(n==-1)
printf("THe queue is empty");
else
printf("The value dequed is %d\n",n);
display(q1);
}
else
break;
}
return 0;
}

Comments

Popular posts from this blog

Introduction to AVL tree

AVL tree is a balanced binary search tree where the difference between heights of two sub trees is maximum 1. Why balanced tree A binary tree is good data structure because search operation here is of the order of O(logn). But this is true if the tree is balanced - which means the left and right subtrees are almost equal in height. If not balanced, search operation will take longer.  In worst case, if the tree has only one branch, then search is of the order O(n). Look at this example.  Here all nodes have only right children.  To search a value in this tree, we need upto 7 iterations, which is O(n). So this tree is very very inefficient. One way of making the tree efficient is to, balance the tree and make sure that height of two branches of each node are almost equal. Height of a node Height of a node is the distance between the node and its extreme child.  In the above a diagram, height of 37 is 3 and height of left child of 37 is 0 and right child of 37 is 2. Bal...

Reverse a singly linked list

One of the commonly used interview question is - how do you reverse a linked list? If you talk about a recursive function to print the list in reverse order, you are so wrong. The question is to reverse the nodes of list. Not print the nodes in reverse order. So how do you go about reversing the nodes. You need to take each node and link it to previous node. But a singly linked list does not have previous pointer. So if n1 is current node, n2 = n1->next, you should set     n2->next = NULL But doing this would cut off the list at n2. So the solution is recursion. That is to reverse n nodes  n1,n2,n3... of a list, reverse the sub list from n2,n3,n4.... link n2->next to n1 set n1->next to NULL The last step is necessary because, once we reverse the list, first node must become last node and should be pointing to NULL. But now the difficulty is regarding the head? Where is head and how do we set it? Once we reach end of list  viz n1->next ==NULL, th...

Balanced brackets

Have you observed something? When ever you are writing code using any IDE, if you write mismatched brackets, immediately an error is shown by IDE. So how does IDE  know if an expression is having balanced brackets? For that, the expression must have equal number of opening and closing brackets of matching types and also they must be in correct order. Let us look at some examples (a+b)*c+d*{e+(f*g)}   - balanced (p+q*[r+u )] - unbalanced (p+q+r+s) ) - unbalanced (m+n*[p+q]+{g+h}) - balanced So we do we write a program to check if an expression is having balanced brackets? We do need to make use of stack to store the brackets. The algorithm is as follows Scan a character - ch from the expression If the character is opening bracket, push it to stack If the character is closing bracket pop a character from stack If popped opening bracket and ch are not of same type ( ( and ) or [ and ] ) stop the function and return false Repeat steps 2 and 3 till all characters are scanned. On...