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1 Scheme of valuvation -1I Subject & Code : Data Structure and Application (15CS33) NOTE: ANSWER All FIVE QUESTIONS 1 Explain the diadvantage of linear queue and how it i olved in circular queue. Explain with uitable example how you would implement circular queue uing dynamically allocated Array. In a normal Queue Data Structure, we can inert element until queue become full. But once if queue become full, we can not inert the next element until all the element are deleted from the queue. For example conider the queue below... 4 After inerting all the element into the queue. Now conider the following ituation after deleting three element from the queue Thi ituation alo ay that Queue i Full and we can not inert the new element becaue, 'rear'' i till at lat poition. In above ituation, even though we have empty poition in the queue we can not make ue of them to inert new element. Thi i the major problem in normal queue data tructure. To overcomee thi problem we ue circular queue data tructure. Algorithm for dynamic allocation of memory What i a Sytem tack? How the control i tranferred to or from the function with the help of activation record. Modern imperative programmingg language typically have local variable. Created upon entry to function. Detroyed when function return. Each invocation of a function ha it own intantiation of local variable. Recurive call to a function require everal intantiation to exit imultaneouly. Function return only after all function it call have returned lat-in-firt-out (LIFO) behavior. A LIFO tructure called a tack i ued to hold each intantiation. The portion of the tack ued for an invocation of a function i called the function tack frame or activation record. What i linked lit? Explain the different type of Linked Lit with example. Linear Linked Lit or One Way Lit or Singly Linked Lit:- It i linear collection of data element which are called Node. The element may or may not be tored in conecutive memory location. So pointer are ued maintain linear order. Each node i divided into two part. The firt part contain the information of the element and i called INFO Field. The econd part contain the addre of the next node and i called LINK Field or NEXT Pointer Field. The START contain the tarting addre of the linked lit i.e. it contain the addre of the firt node of the linked lit. The LINK Field of lat node contain NULL Value whichh indicate that it i the end of linked lit. It i hown 4 below:

2 2. Doubly Linked Lit or Two-Way Linked Lit or Two-Way Chain:- In it each node i divided into three part: 1. The firt part i PREV part. It i previou pointer field. It contain the addre of the node which i before the current node. 2. The econd part i the INFO part. It contain the information of the element. 3. The third part i NEXT part. It i next pointer field. It contain the addre of the node which i after the current node. There are two pointer FIRST and LAST. FIRST point to the firt node in the lit. LAST point to the lat node in the lit. The PREV field of firt node and the NEXT field of lat node contain NULL value. Thi how the end of lit on both ide. Thi lit can be travered in both direction that i forward and backward. It i hown below: 3. Circular Linked Lit: 1.SLL 2.DLL In it the lat node doe not contain NULL pointer. Intead the lat node contain a pointer that ha the addre of firt node and thu point back to the firt node. It i hown below: 4 Write a C-Program to implement the inert, delete and diplay operation on tack uing Linked Lit. nclude<tdio.h> #include<conio.h> truct Node int data; truct Node *next; *top = NULL; void puh(int); void pop(); void diplay(); void main()

3 int choice, value; clrcr(); printf("\n:: Stack uing Linked Lit ::\n"); while(1) printf("\n****** MENU ******\n"); printf("1. Puh\n2. Pop\n3. Diplay\n4. Exit\n"); printf("enter your choice: "); canf("%d",&choice); witch(choice) cae 1: printf("enter the value to be inert: "); canf("%d", &value); puh(value); break; cae 2: pop(); break; cae 3: diplay(); break; cae 4: exit(0); default: printf("\nwrong election!!! Pleae try again!!!\n"); void puh(int value) truct Node *newnode; newnode = (truct Node*)malloc(izeof(truct Node)); newnode->data = value; if(top == NULL) newnode->next = NULL; ele newnode->next = top; top = newnode; printf("\ninertion i Succe!!!\n"); void pop() if(top == NULL) printf("\nstack i Empty!!!\n"); ele truct Node *temp = top; printf("\ndeleted element: %d", temp->data); top = temp->next; free(temp); void diplay() if(top == NULL) printf("\nstack i Empty!!!\n"); ele truct Node *temp = top; while(temp->next!= NULL) printf("%d--->",temp->data); temp = temp -> next; printf("%d--->null",temp->data);

4 5 What i recurion?write a recurive function for computing the N th term of a Fibonacci equence.alo give the trace of tack content for N=3. Fibonacci Serie uing Recurion #include<tdio.h> int fib(int n) if (n <= 1) return n; return fib(n-1) + fib(n-2); int main () int n = 9; printf("%d", fib(n)); getchar(); return 0; fib(5) / fib(4) fib(3) / / fib(3) fib(2) fib(2) fib(1) / / / fib(2) fib(1) fib(1) fib(0) fib(1) fib(0) / fib(1) fib(0) 6 Write a recurive algorithm to erach an element uing binary erach.ue thi algorithm to find the element 3 in the array 1,3,7,15,32,36,7,95,10.Show the tracing of algorithm. nt binarysearch(int arr[], int l, int r, int x) if (r >= l) int mid = l + (r - l)/2; // If the element i preent at the middle itelf if (arr[mid] == x) return mid; // If element i maller than mid, then it can only be preent // in left ubarray if (arr[mid] > x) return binarysearch(arr, l, mid-1, x); // Ele the element can only be preent in right ubarray return binarysearch(arr, mid+1, r, x); 7 // We reach here when element i not preent in array return -1; Write a function to 1.Concatenate 2 Singly Linked Lit 2.To create Ordered Linked Lit node * concatenate (node *head1, node *head2) node *p; if (head1==null) //if the firt linked lit i empty return (head2);

5 if (head2==null) lit i empty return (head1); p=head1; node of the firt linked lit //if econd linked //place p on the firt while (p->next!=null) //move p to the lat node p=p->next; p->next=head2; //addre of the firt node of the econd linked lit tored in the lat node of the firt linked lit return(head1); (b)void ortedinert(truct Node** head_ref, truct Node* new_node) truct Node* current; /* Special cae for the head end */ if (*head_ref == NULL (*head_ref)->data >= new_node->data) new_node->next = *head_ref; *head_ref = new_node; ele /* Locate the node before the point of inertion */ current = *head_ref; while (current->next!=null && current->next->data < new_node->data) current = current->next; new_node->next = current->next; current->next = new_node; 2+ 6 What i Priority Queue? Write an function to inert an element into Priority queue uing array. In computer cience, a priority queue i an abtract data type which i like a regularqueue or tack data tructure, but where additionally each element ha a "priority" aociated with it. In a priority queue, an element with high priority i erved before an element with low priority. id add ( truct pque *pq, truct data dt ) truct data temp ; int i, j ; if ( pq -> rear == MAX - 1 ) printf ( "\nqueue i full." ) ; return ; pq -> rear++ ; pq -> d[pq -> rear] = dt ; if ( pq -> front == -1 ) pq -> front = 0 ; for ( i = pq -> front ; i <= pq -> rear ; i++ )

6 for ( j = i + 1 ; j <= pq -> rear ; j++ ) if ( pq -> d[i].prno > pq -> d[j].prno ) temp = pq -> d[i] ; pq -> d[i] = pq -> d[j] ; pq -> d[j] = temp ; ele if ( pq -> d[i].prno == pq -> d[j].prno ) if ( pq -> d[i].ord > pq -> d[j].ord ) temp = pq -> d[i] ; pq -> d[i] = pq -> d[j] ; pq -> d[j] = temp ; 9 10 A circulat queue Q the ize of which i 5 ha three element 20,60,50 where front=0 and rear=2 Show with neceary diagram, what i the value of front and rear after each of thee operation. 1.Inert item 50 2.inert item 10 3.Inert item 30 4.Delete an item Explain with diagram Write a Program in C to create a linked lit and interchange the element to the lit at poition M and N and diplay the content of the lit before and after interchanging the element. Void wap(node front,int m,int n) NODE temp=front; NODE m, n, tr; Int i; For(i=1;i<m;i++) Temp=temp-<next; M=temp; Temp=front; For(i=1;i<=n;i++) Temp=temp->next N=temp; Temp=front; For(i=1;i<n;i1++) Tr=(temp->next)->next; Temp->next=m; Temp->next=tr; Temp=front; For(i=1;i<m;i++) Tr=temp->next-next; Temp->next=n; 4*2

7 Temp->next=tr; *******

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