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DATA STRUCTURES ( C++ ) This PPT is Dedicated to my inner controller AMMA BHAGAVAN-Oneness Founders. Developed by,EDITED BY, S.V.G.REDDY,M.Siva Naga Prasad,

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Presentation on theme: "DATA STRUCTURES ( C++ ) This PPT is Dedicated to my inner controller AMMA BHAGAVAN-Oneness Founders. Developed by,EDITED BY, S.V.G.REDDY,M.Siva Naga Prasad,"— Presentation transcript:

1 DATA STRUCTURES ( C++ ) This PPT is Dedicated to my inner controller AMMA BHAGAVAN-Oneness Founders. Developed by,EDITED BY, S.V.G.REDDY,M.Siva Naga Prasad, Associate professor, student of M.tech(SE) Dept.of CSE,GIT, GITAM UNIVERSITY

2 STACK USING ARRAYS Let us take an array a[5] and take a variable top points to -1. PUSH:  To INSERT the element in to stack using top.  Here we check for the OVERFLOW condition. POP:  To RETRIEVE elements from the stack using top.  Here we check for the UNDERFLOW condition. This concept is nothing but LIFO. SOURCE CODE: /* Program To Implement Stack using Array */ #include #define MAX 10

3 class stack { private : int sp, a [ MAX ]; public : void init ( ); void push ( int ); void pop ( ); void display ( ); void count ( ); }; void stack :: init ( ) { sp = - 1; } void stack :: push ( int data) { if (sp = = ( MAX – 1 ) ) { cout<<"\n STACK OVERFLOW \n"; return; }

4 sp + + ; a [ sp ] = data; } void stack :: pop ( ) { if ( sp < 0 ) { cout<<"\n STACK UNDERFLOW.....\n"; return; } cout<<"\n POPED DATA IS ::: "< = 0 ; i - -) cout << a [ i ] <<"\t"; } void stack :: count ( ) { cout<<"\n NUMBER OF ELEMENTS IN A STACK ARE ::: "<<(sp+1); };

5 void main ( ) { stack ob; int data,ch; clrscr ( ); ob.init ( ); cout<<"\n**********STACK OPERATIONS**********\n"; cout<<"\n1.Push Operation"; cout<<"\n2.Pop Operation"; cout<<"\n3.Display Operation"; cout<<"\n4.Count Operation"; cout<<"\n5.Exit Operation"; cout<<"\n*************************************\n"; do { cout<<"\n ENTER YOUR CHOICE :: "; cin>>ch; switch ( ch ) { case 1:cout<<"\n ENTER ELEMENT TO BE INSERTED :::"; cin>>data; ob.push ( data ); break; case 2: ob.pop ( ); break; case 3: ob.display ( ); break; case 4:ob.count ( );break; case 5: exit ( 0 ); defualt: cout<<"\nINVALID CHOICE "; } while ( ch ! = 5 ); getch ( ); }

6 OUTPUT:

7 STACK USING LINKED LIST We will create a linked list and insert an element ‘10’ and address as ‘0’.using top for the first node. For second node insert data element ‘20’ and insert first node address at second node address field. For third node insert data element ‘30’ and insert second node address at third node address field. after thirty we will stop the process. If we want to print the elements 30,20,10 will be displayed, Thiss follows LIFO conceot.

8 Source code: #include class st { public: struct node { int data; struct node *next; }*start,*temp,*top; st() { start=temp=top=NULL; } void create() { int d; cout<<"Enter data"; cin>>d; if(start==NULL) { start=new node; start->data=d;

9 start->next=NULL; top=start; } else { temp=new node; temp->data=d; temp->next=top; top=temp; } void disp() { while(top!=NULL) { cout data<<"\t"; top=top->next; } }; void main() { st ob; int ch; clrscr();

10 while(ch) { cout<<"Enter ur choice"; cout<<"0 STOP\n1 CREATE\n 2 READ"; cin>>ch; if(ch==1) ob.create(); else if(ch==2) ob.disp(); } OUTPUT:

11 QUEUE USING ARRAYS Here we will take an array a[5],and two variables front, rear points to -1. WRITE:  Here will insert the element into the queue using rear variable.  Here check for the Overflow condition. READ:  Here we will retrieve the elements from the queue using front variable.  Here check for the Underflow condition. This follows the FIFO concept rear front

12 SOURCE CODE: /* Program To Implement Queue using Array */ #include #define MAX 10 class queue { private : int front, rear, a [ MAX ]; public : void init ( ); void write ( int ); void read ( ); void count ( ); void display ( ); }; void queue :: init ( ) { front = rear = - 1; } void queue :: write ( int data) { if ( rear = = ( MAX - 1 ) ) cout<<"\n QUEUE IS OVERFLOW......"; else a [ + + rear ] = data; } void queue :: read ( ) { if( front = = rear ) cout<<"\n QUEUE IS UNDERFLOW....."; else cout<<"\n DELETED ELEMENT IN QUEUE IS :: "<

13 void queue :: count ( ) { cout<<"\n NUMBER OF ELEMENTS IN A QUEUE ARE :: "<<(rear-front); } void queue :: display ( ) { cout<<"\n ELEMENTS IN A QUEUE ARE:: "; for( int i = (front + 1); i < = rear; i + + ) cout<< a [ i ]<<"\t"; } void main ( ) { queue ob; int ch,data; clrscr ( ); ob.init ( ); cout<<"\n*****QUEUE OPERATIONS****\n"; cout<<"\n1.Write "; cout<<"\n2.Read "; cout<<"\n3.Count"; cout<<"\n4.Display"; cout<<"\n5.Exit"; cout<<"**************************\n";

14 do { cout<<"\n ENTER YOUR CHOICE :: "; cin>>ch; switch ( ch ) { case 1:cout<<"\n ENTER ELEMENT TO BE INSERTED IN QUEUE :: "; cin>>data; ob.write ( data ); break; case 2:ob.read ( ); break; case 3:ob.count ( ); break; case 4:ob.display ( ); break; case 5:exit ( 0 ); break; default :cout<<"\n INVALID CHOICE..."; } while( ch ! = 5 ); getch ( ); }

15 OUTPUT:

16 Queue using linked list Here we will create linked list with ‘n’ nodes one after another 10,20,30 etc. If we try to print the elements it will display as 10,20,30. which follows FIFO concept.

17 SOURCE CODE: /* Program To Implement Queue using Linked List */ #include #define NULL 0 class node { int data; node *next; public: void create ( node *); void print ( node *); }; void node :: create (node *list) { cout<<"\n ENTER THE INPUT NO :: "; cout<<"\n TYPE 999 AT THE END :: "; cin>>list->data; if(list -> data = = 999) list->next = NULL; else { list -> next = new node; create( list -> next); } return; }

18 void node :: print (node *list) { if( list -> next ! = 0) { cout data; cout "; } else return; print( list -> next); } void main ( ) { node *head, ob; clrscr ( ); head = new node; ob.create ( head ); cout<<"\n QUEUE ELEMENTS ARE:: "; ob.print( head ); cout<<"999"; getch ( ); }

19 OUTPUT:

20 BINARY TREE USING RECURSION A binary tree is a tree data structure in which each node has at most two children. Typically the child nodes are called left and right. Binary trees are commonly used to implement binary search trees and binary heaps. Starting at the root of a binary tree, there are three main steps that can be performed and the order in which they are performed define the traversal type. There are 3 types of traversals:  1. Pre-Order  2. In-Order  3. Post-Order To traverse a non-empty binary tree in preorder, perform the following operations recursively at each node, starting with the root node: 1.Visit the root. 2.Traverse the left sub tree. 3.Traverse the right sub tree. To traverse a non-empty binary tree in in order, perform the following operations recursively at each node, starting with the root node: 1.Traverse the left sub tree. 2.Visit the root. 3.Traverse the right sub tree. To traverse a non-empty binary tree in post order, perform the following operations recursively at each node, starting with the root node: 1.Traverse the left sub tree. 2.Traverse the right sub tree. 3.Visit the root.

21 BINARY TREE: Preorder:- 15,7,22 will be displayed. Post order:- 7,22,15 will be displayed. In order:- 7,15,22 will be displayed.

22 SOURCE CODE: /* Program To Implement Binary Tree Traversing */ #include class bstree { public: struct node { int data; node *left; node *right; }*head; void create (node *); void inorder (node *); void preorder (node *); void postorder (node *); }; void* bstree:: create(node *list) { node *temp1,*temp2; int val; if(list = = NULL) { list = new node; cout<<"\nEnter Data Element:: "; cin>>list->data; list -> left = list -> right = NULL; } else

23 { cout<<"\n enter the data element"; cin>>val; temp1 = list; while( temp1 ! = NULL ) { temp2 = temp1; if(temp1 -> data > val) temp1 = temp1 -> left; else temp1 = temp1 -> right; } if(temp2 -> data > val) { temp2 -> left = new node; temp2 = temp2 -> left; temp2 -> data = val; temp2 -> left = temp2 -> right = NULL; } else { temp2 -> right = new node; temp2 = temp2 -> right; temp2 -> data = val; temp2 -> left = temp2 -> right = NULL; } return (list); }

24 void bstree:: inorder(node *root) { if( ! root ) return; inorder( root -> left ); cout data<<"\t"; inorder( root -> right ); } void bstree::preorder(node*root) { if( ! root ) return; cout data<<”\t”; preorder( root -> left ); preorder( root -> right); } void bstree::postorder(node*root) { if( ! root) return; postorder( root -> left ); postorder( root -> right ); cout data<<”\t”; }

25 void main ( ) { node n,*head; head = NULL; clrscr ( ); cout<<"\nCreate A Binary Tree\n"; head=n.create ( head ); cout<<"\n the inorder traversal gives the following nodes"; n.inorder ( head ); getch ( ); } OUTPUT:

26

27 BINARY SEARCH TREE  A tree having left child less than parent and right child grater than the parent.  Traversals are same as binary tree.

28 SOURCE CODE: /* Program to implement Binary search tree */ #include class btree { private : struct btreenode { btreenode *leftchild ; int data ; btreenode *rightchild ; } *root; public: btree ( ) ; void buildtree ( int num ) ; static void insert ( btreenode **sr, int num ) ; void traverse ( ) ; static void inorder ( btreenode *sr ) ; static void preorder ( btreenode *sr ) ; static void postorder ( btreenode *sr ) ; static void del ( btreenode *sr ) ; ~btree ( ) ; } ;

29 btree :: btree ( ) { root = NULL ; } void btree :: buildtree ( int num ) { insert ( &root, num ) ; } void btree :: insert ( btreenode **sr, int num ) { if ( *sr == NULL ) { *sr = new btreenode ; ( *sr ) -> leftchild = NULL ; ( *sr ) -> data = num ; ( *sr ) -> rightchild = NULL ; return ; } else // search the node to which new node will be attached { // if new data is less, traverse to left if ( num data ) insert ( & ( ( *sr ) -> leftchild ), num ) ; else // else traverse to right insert ( & ( ( *sr ) -> rightchild ), num ) ; } return ; }

30 void btree :: traverse( ) { cout << "\nIN - ORDER TRAVERSAL :: " ; inorder ( root ) ; cout << "\nPRE - ORDER TRAVERSAL :: " ; preorder ( root ) ; cout << "\nPOST - ORDER TRAVERSAL :: " ; postorder ( root ) ; } void btree :: inorder ( btreenode *sr ) { if ( sr != NULL ) { inorder ( sr -> leftchild ) ; cout data ; inorder ( sr -> rightchild ) ; } else return ; } void btree :: preorder ( btreenode *sr ) { if ( sr != NULL ) { // print the data of a node cout data ; // traverse till leftchild is not NULL preorder ( sr -> leftchild ) ; // traverse till rightchild is not NULL preorder ( sr -> rightchild ) ; }

31 else return ; } void btree :: postorder ( btreenode *sr ) { if ( sr != NULL ) { postorder ( sr -> leftchild ) ; postorder ( sr -> rightchild ) ; cout data ; } else return ; } btree :: ~btree( ) { del ( root ) ; } void btree :: del ( btreenode *sr ) { if ( sr != NULL ) { del ( sr -> leftchild ) ; del ( sr -> rightchild ) ; } delete sr ; }

32 void main( ) { btree bt ; int req, i = 1, num ; clrscr(); cout << "\n SPECIFY THE NUMBER OF ITEMS TO BE INSERTED :: " ; cin >> req ; while ( i + + <= req ) { cout << "\n ENTER THE DATA :: " ; cin >> num ; bt.buildtree ( num ) ; } bt.traverse( ) ; getch(); } OUTPUT:

33 SPARSE MATRIX AIM: Write a program in C++ to implement ADDITION and MULTIPLICTION of two SPARSE matrixes. THEORY: If a lot of elements from a matrix have a value 0 then the matrix is known as SPARSE MATRIX. If the matrix is sparse we must consider an alternate way of representing it rather the normal row major or column major arrangement. This is because if majority of elements of the matrix are 0 then an alternative through which we can store only the non-zero elements and keep intact the functionality of the matrix can save a lot of memory space. Example: Sparse matrix of dimension 7 x 7. COLUMNS ROWS

34 A common way of representing non-zero elements of a sparse matrix is the 3-tuple forms. In this form each non-zero element is stored in a row, with the 1 st and 2 nd element of this row containing the row and column in which the element is present in the original matrix. The 3 rd element in this row stores the actual value of the non- store element. For example 3-tuple representation of the sparse matrix as shown in below. int spmat[10][3]={ 7,7,9, 0,3,-5, 1,1,4, 1,6,7, 2,4,9, 3,1,3, 3,3,2, 4,0,11, 4,2,2, 6,2,8 }

35 SOURCE CODE: /*Program to demonstrate addition and multiplication of Two Sparse Matrix */ #include #define x 25 class sparce { private: int a [ x ] [ x ], b [ x ] [ x ], c [ x ] [ x ], m, n, p, q; public: void init ( ); void input ( ); void add ( ); void mul ( ); void display ( int [25][25], int, int ); void convert( int [25][25], int, int ); }; void sparce :: init ( ) { int i, j; for(i = 0; i < x;i + + ) for( j = 0; j < x; j + +) c [ i ] [ j ] = 0; }

36 void sparce :: input() { int i,j; cout<<"\nEnter order Of First matrix::"; cin>>m>>n; cout<<"\nEnter order Of Second matrix::"; cin>>p>>q; cout<<"\nEnter"<> a[ i ] [ j ]; cout<<"\nEnter"<>b [ i ] [ j ]; } void sparce :: add ( ) { int i, j; if( m = = p && n = = q ) { for( i = 0 ; i < m ; i + + ) for( j = 0; j < n; j + + ) c[ i ] [ j ] = a [ i ][ j ] + b [ i ] [ j ]; convert( c, m, n); } else cout<<"\nAddition Is Not Possible"; }

37 void sparce :: mul ( ) { int i, j, k; if(n = = p) { for( i = 0; i < m; i + +) for( j = 0; j < q; j + + ) for( k = 0; k < n; k + + ) c[ I ] [ j ] + = a [ I ] [ k ] * b [ k ] [ j ]; convert(c, m, n); } else cout<<"\n Multiplecation Is Not Possible"; } void sparce :: display(int c[25][25], int m, int n) { int i,j; for( i = 0 ;i < m; i + + ) { for( j = 0 ; j < n ; j + + ) cout<

38 void sparce :: convert(int c[25][25], int m, int n) { int i, j, k = 1,t = 0; int sp[25][25]; for( i = 0 ; i < m ; i + +) for( j = 0 ; j < n ; j + + ) if(c [ i ] [ j ] ! = 0 ) { sp [ k ] [ 0 ] = i; sp [ k ] [ 1 ] = j; sp [ k ] [ 2 ] = c [ i ] [ j ]; k + + ; t + + ; } sp[ 0 ] [ 0 ] = m; sp[ 0 ] [ 1 ] = n; sp[ 0 ] [ 2 ] = t; display( sp, k, 3); } void main ( ) { sparce ob; clrscr ( ); ob.init ( ); ob.input ( ); cout<<"\nAddition of Two Sparce Matrix\n"; ob.add ( ); ob.init ( ); cout<<"\nMultiplecation Of Two Sparce Matrix\n"; ob.mul ( ); getch ( ); }

39 OUTPUT:

40 INFIX TO POSTFIX CONVERTIONic Suppose Q is an arithmetic expression written in infix notation. This algorithm finds the equivalent postfix expression P. Step 1. Push “(“ onto stack and add “)” to the end of Q. 2. Scan Q from left to right and repeat step 3 to 6 for each element of Q until the stack is empty. 3. If an operand is encountered, add it to p. 4. If a left parenthesis is encountered,push it onto stack. 5 If an operator * is encountered, then: a. repeatedly pop from stack and top each operator (on the top of stack ) which has the same precedence or higher precedence than *. b. Add * to stack. 6. If a right parenthesis is encountered, then: a. repeatedly from stack and add to P each operator (on the top of stack) until a left parenthesis is encountered. b. remove the left parenthesis [ Do not add the left parenthesis top] [End of if structure] [End of step 2 loop] 7. Exit

41 (A+(B*C-(D/E^F)*G)*H) Symbol scanned stack Expression P 1 A(A 2 +( +A 3 (( + (A 4 B( + (AB 5 *( + ( *AB 6 C( + ( *ABC 7 -( + ( -ABC* 8 (( + ( - (ABC* 9 D( + ( - (ABC*D 10 /( + ( - ( /ABC*D 11 E( + ( - ( /ABC*DE 12 ^( + ( - ( / ^ABC*DE 13 F( + ( - ( / ^ABC*DEF 14 )( + ( -ABC*DEF^/ 15 *( + ( - *ABC*DEF^/ 16 G( + ( - *ABC*DEF^/G 17 )( +ABC*DEF^/G*- 18 *( + *ABC*DEF^/G*- 19 H( + *ABC*DEF^/G*-H 20 )ABC*DEF^/G*-H

42 SOURCE CODE: /* Program To implement infix to postfix Expression */ #include char stack[30], postfix[30], infix[30]; int top = - 1; int pri( char x ) { int value; switch ( x ) { case ')': value=0; break; case '+': case '-': value=1; break; case '*': case '/': case '%': value=2;break; case '^': value=3;break; case '(': value=4;break; default: cout<<"INVALID EXPRESSION !!!!!!"; exit(1); } return value; }

43 void push ( char x ) { top = top + 1; stack [top] = x; } char stacktop ( ) { return stack [ top ]; } int isalnum (char x) { return ( (x>='0' && x ='a' && x ='A' && x<='Z')); } char pop( ) { return stack[top - - ]; }

44 void intopost(char infix[ ], char postfix[ ]) { int i, j=0; char c, pc; for ( i = 0; ( c = infix[ i ] ) != '\0' ; i + +) { if ( isalnum (c) ) postfix [ j + + ] = c; else { while ( top ! = - 1 && (pri (stacktop () ) >= pri (c) ) ) { If ( stacktop( ) = = '(' && c! = ')' ) break; if ( stacktop( ) = = '(' && c = =')' ) { pop () ; break; } pc = pop( ); if ( pc! = '(' ) postfix [ j + + ] = pc; else break; } if( c! = ')' ) push ( c ); } while( top ! = -1 ) postfix[ j + + ] = pop( ); postfix [ j ] = '\0'; }

45 void main ( ) { clrscr ( ); cout<<"ENTER INFIX EXPRESSION ::\n\n\t\t\t"; cin>>infix; intopost( infix, postfix ); cout<<"POSTFIX EXPRESSION ::\n\n\t\t\t "; cout<

46 POSTFIX EVALUATION THEORY: Reverse Polish notation is a mathematical notation wherein every operator follows all of its operands. It is also known as Postfix notation and is parenthesis free. In Reverse Polish notation the operators follow their operands; for instance, to add three and four, one would write “3 4 +” rather than “3 + 4”. If there are multiple operations, the operator is given immediately after its second operand; so the expression written “3 − 4 + 5” in conventional infix notation would be written “3 4 − 5 +” in RPN: first subtract 4 from 3, then add 5 to that. Infix Expression: Any expression in the standard form like "2*3-4/5" is an Infix(In order) expression. Postfix Expression: The Postfix(Post order) form of the above expression is "23*45/-". Postfix Evaluation: In normal algebra we use the infix notation like a+b*c. The corresponding postfix notation is abc*+. The algorithm for the conversion is as follows: Scan the Postfix string from left to right. Initialize an empty stack. If the scanned character is an operand, add it to the stack. If the scanned character is an operator, there will be at least two operands in the stack  If the scanned character is an Operator, then we store the top most element of the stack(topStack) in a variable temp. Pop the stack. Now evaluate topStack(Operator)temp. Let the result of this operation be retVal. Pop the stack and Push retVal into the stack.  Repeat this step till all the characters are scanned. After all characters are scanned, we will have only one element in the stack. Return topStack.

47 StackExpression Example: Postfix String: * Initially the Stack is empty. Now, the first three characters scanned are 1,2 and 3, which are operands. Thus they will be pushed into the stack in that order. Next character scanned is "*", which is an operator. Thus, we pop the top two elements from the stack and perform the "*" operation with the two operands. The second operand will be the first element that is popped. The value of the expression(2*3) that has been evaluated(6) is pushed into the stack. StackExpression StackExpression

48 StackExpression Next character scanned is "+", which is an operator. Thus, we pop the top two elements from the stack and perform the "+" operation with the two operands. The second operand will be the first element that is popped. The value of the expression(1+6) that has been evaluated(7) is pushed into the stack. Next character scanned is "4", which is added to the stack. Next character scanned is "-", which is an operator. Thus, we pop the top two elements from the stack and perform the "-" operation with the two operands. The second operand will be the first element that is popped. StackExpression StackExpression

49 StackExpression The value of the expression(7-4) that has been evaluated(3) is pushed into the stack. Now, since all the characters are scanned, the remaining element in the stack (there will be only one element in the stack) will be returned. End result: Postfix String : * Result : 3 StackExpression

50 SOURCE CODE: /*Program To Evaluate Postfix Expression */ #include class postfix { private: int stack[50], len, top; char post[50]; public: postfix ( ); void push ( int ); int pop ( ); int pfix ( ); }; void postfix :: postfix ( ) { top = - 1; } int postfix :: pfix ( ) { int a, b, i, temp; cout<<"\nEnter Postfix Expression::"; cin>>post; len = strlen ( post ); post [ len] = '#';

51 for( i = 0 ; post [ i ] ! = '#' ; i + +) { if( post [ i ] = '0') push( post [ i ] - 48); else { a = pop ( ); b = pop ( ); switch ( post [ i ]) { case '+': temp = b + a;break; case '-': temp = b - a; break; case '*': temp = b * a; break; case '/': temp = b/a; break; case '%': temp = b%a; break; case '^': temp = pow( b, a ); } push ( temp ); } return( pop ( ) ); }

52 void postfix :: push( int x ) { stack[ + + top ] = x; } int postfix :: pop ( ) { int x = stack [ top ]; top- -; return x; } void main ( ) { int x; postfix ob; clrscr ( ); x=ob.pfix ( ); cout<<"\nResult Of Postfix Expression Is\t"<

53 Quick Sort right 1)When pivot is at left end, 1)Compare a[pivot] with a[right] element if (a[pivot] < a[right]) then right-- else swap a[pivot] and a[right] 2)When pivot is at right end, Compare a[pivot] with a[left] element if (a[left] < a[pivot]) then left++ else swap a[left] and a[pivot] pivot left

54 STEP1: left, pivot right STEP2: left right, pivot STEP3: left right, pivot STEP4: left right, pivot STEP5: left, pivot right STEP6: left, pivot right

55 STEP7: left, pivot right STEP8: left, pivot right STEP9: left right, pivot STEP10: left, right, pivot Here we will stop the main process as the left and right pointers are equal. Now see the elements left to ‘11’ are less than ‘11’ and elements right to ‘11’ are grater than ‘11’. Now divide the main list into 2 sub lists such as(2,7,3) and (46,89,21,34) and do the same above process.

56 SOURCE CODE: /*Program To Implement Quick Sort */ #include class qsort { private : int n; public : void sort(int *, int, int); void swap(int *, int *); }; void qsort::sort( int *a, int l, int u ) { int i, j, pivot, k; if( l < u ) { i = pivot = l; j=u; while ( i < j ) { while( a [ i ] <= a [ pivot ] ) i + +; while( a[ j ] > a [ pivot ] ) j - -; if ( i < j ) swap( &a [ i ], &a [ j ] ); }

57 sort( a, l, j-1 ); sort( a, j+1, u ); } void qsort::swap(int *x, int *y) { int temp; temp = *x; *x = *y; *y = temp; } void main ( ) { int a[50],i, n; qsort obj; clrscr ( ); cout<<"\n ENTER THE NUMBER OF ELEMENTS OF ARRAY :: "; cin>>n; cout<<"\n ENTER THE ELEMENTS :: \n \t"; for( i = 0; i >a[ i ]; cout<<"\n ELEMENTS BEFORE SORTING ARE ::\n \t "; for(i=0;i

58 obj.sort(a,0,n-1); cout<<"\n ELEMENTS AFTER SORTING ARE :: \n \t "; for( i = 0 ; i < n ; i + + ) cout<

59 Selection sort Consider the elements as shown, min i Here min is compared with a[1]  as min is > a[1]  min=a[1] min i This min is compared with a[2],as this is < a[2]  min is same that is 33 This min is compared with a[3],as this is > a[3]  min =a[3]. min I Now this is compared with a[4],a[5],a[6],a[7] as min is less than all of these min remains 33 At last swap min and a[i] like this continue the process with i=1,2,3……

60 SOURCE CODE: #include class selsort {public : void sort(int *, int); }; void selsort::sort(int *a, int n) { int i, j, x, min, temp; for( i = 0 ; i < ( n – 1 ) ; i + + ) { x = i; min = a [ i ]; for( j = i + 1; j < n; j + + ) { if( min > a [ j ] ) { min = a [ j ]; x = j; } temp = a [ i ] ;a [ i ] = a [ x ]; a [ x ] = temp; } void main( ) { int a[50], n, i; clrscr( );

61 cout<<"\n ENTER THE SIZE OF THE ARRAY: \n\t "; cin>>n; cout<<"\n ENTER THE ELEMENTS:\n\t"; for( i = 0 ;i < n ;i + + ) cin>>a [ i ]; cout<<"\n ELEMENTS BEFORE SORTING:\n\t"; for( i = 0 ; i < n ; i + + ) cout<

62 LINEAR SEARCH Here we want to search for ‘50’. So compare ’50’ with a[i] where i=0,1,2,3,….  If (a[i]==50) Then element is found at location i that is 4 Else i++ Here the time complexity is O(n).

63 SOURCE CODE: #include class lsearch { private: int a[50], n, count, key; public: void init ( ); void linear ( ); }; void lsearch::init ( ) { count = 0; } void lsearch::linear ( ) { int i; clrscr ( ); cout<<"\nENTER SIZE OF AN ARRAY :: "; cin>>n; cout<<"\n\nENTER "<> a [ i ]; cout<<"\n\nENTER SEARCH ELEMENT :: "; cin>>key; cout<<"\n\nELEMENTS IN ARRAY ARE :\n";

64 for( i = 0; i < n; i + +) cout<< a [ i ]<<"\t"; for( i = 0; i < n; i + + ) if(a [ i ] = = key) { count + +; break; } if( count = = 1 ) cout<<"\n\n ELEMENT IS FOUND IN "<< ( i + 1)<<" LOCATION"; else cout<<"\nELEMENT IS NOT FOUND...."; } void main ( ) { lsearch ob; clrscr ( ); ob.init (); ob.linear ( ); getch ( ); }

65 OUTPUT:

66 BINARY SEARCH Here elements must be in Ascending/Descending order. Consider the elements in ascending order low high here low=0 and high=7 Then calculate mid=(low+high)/2 Let us search for k=71 If (a[mid]==k) then element is found at ‘mid’ location If(k

67 SOURCE CODE: /*Program To Implement Binary Search */ #include class bsearch { private : int a[50], n, x; public : void binary ( ); }; void bsearch::binary ( ) { int i, j, temp, mid, beg, end; beg = 0; cout<<"\n\nENTER THE SIZE OF THE ARRAY :: "; cin>>n; end = n - 1; cout<<"\n\nENTER THE ELEMENTS OF THE ARRAY :: "; for( i = 0; i < n; i + +) cin>>a[i]; cout<<"\n\nELEMENTS BEFORE BEFORE SORTING ARE :: "; for( i = 0; i < n; i + + ) cout<< a [ i ]<<" ";

68 for( i = 0; i < n; i + + ) { for( j = i + 1; j < n; j + +) { if( a[ i ] > a[ j ] ) { temp = a [ i ]; a[ i ] = a[ j ]; a[ j ] = temp; } cout<<"\n\nELEMENTS AFTER SORTING ARE :: "; for( i = 0; i < n; i + + ) cout<>x; while ( beg < = end ) { mid = ( beg + end ) / 2; if ( a [ mid ] = = x ) {

69 cout<<"\nSEARCHING IS SUCCESSFUL AND THE ELEMENTS IS PRESENT AT "<< ( mid + 1 )<<" LOCATION"; return; } else if(x

70 POLINOMIAL ADDITION AND MULTIPLICATION 1 expression: 3x 2 +2x+1 Store all the coefficients 1,2,3 into an array1. 1 expression: 2x 2 +1x+2 Store all the coefficients 2,1,2 into an array2. ADDITION: 3x 2 +2x+1 2x 2 +1x+2 5x 2 +3x+3 Store the result expression coefficients in array3

71 SOURCE CODE: /*Program To Demonstrate Addition And Multiplication Of Two Polynomial Expression */ #include #define n 100 class poly { private: int a[n], b[n], add[n], mul[n], p, q, at; public: void init ( ); void input ( ); void process ( ); void display ( ); }; void poly :: init ( ) { int i; for( i = 0; i < n; i + + ) a[ i ] = b [ i ] = add[ i ] = mul[ i ] = 0; }

72 void poly :: input ( ) { int i; cout<<"\nEnter Degree Of First Polynomial::"; cin>>p; cout<<"\nEnter Degree Of Second Polynomial::"; cin>>q; cout<<"\nEnter Values First Polynomial\n"; for( i = 0; i <= p; i + + ) { cout<<"\nEnter X^"<>a[ i ]; } cout<<"\nEnter Values First Polynomial\n"; for( i = 0; i <= q; i + + ) { cout<<"\nEnter X^"<>b[ i ]; }

73 void poly :: process ( ) { int i, j; if( p > q ) at = p; else at = q; for ( i = 0; i <= at; i + +) add[ i ] = a[ i ] + b[ i ]; for( i = 0; i <= p; i + + ) for( j = 0; j <= q; j + + ) mul [ i + j ] + = a [ i ] * b [ j ]; } void poly :: display ( ) { int i; cout<<"\Addition Of Two Polynomial Expressions Are\n\n"; for( i = at; i >=0 ; i - -) cout< = 0; i - -) cout<

74 void main() { poly ob; clrscr ( ); ob.init ( ); ob.input ( ); ob.process ( ); ob.display ( ); getch ( ); } OUTPUT:

75 SINGLE LINKED LIST THEORY: Figure shows a Linked List. Each item in the list is called a node and contain two fields, a data field and a next address field. The data field holds the actual element on the list. The next address field contains the address of the next node in the list. Such an address which is used to access a particular node, is known as a pointer. The entire linked list is accesses from an external pointer list, that points to the first node in the list. The next field of last node in the list contains a special value, known as NULL. The null pointer is used to signal the end of the list. The singly-linked list is the most basic of all the linked data structures. A singly-linked list is simply a sequence of dynamically allocated objects, each of which refers to its successor in the list. Despite this obvious simplicity, there are myriad implementation variations. The following code inserts a node after an existing node in a singly linked list. The diagram shows how it works. Inserting a node before an existing one cannot be done; instead, you have to locate it while keeping track of the previous node.

76 Similarly, we have functions for removing the node after a given node, and for removing a node from the beginning of the list. The diagram demonstrates the former. To find and remove a particular node, one must again keep track of the previous element.

77 SOURCE CODE: /*Program To Implement Single Linked list */ #include class slist { private: struct list { int data; struct list *next; }*start,*temp,*curr,*add,*tem,*addr; public: void init ( ); void create ( ); void disp ( ); list *search ( int ); void insert ( ); void del ( ); };

78 void slist :: init ( ) { start = temp = curr = NULL; } void slist::create ( ) { char ch; temp = new list; cout<<"\n ENTER THE DATA TO BE STORED \n"; cin>> temp->data; temp->next = NULL; start = curr = temp; cout<<"\n DO YOU WANT TO INSERT ANOTHER NODE (Y/N)"; cin>>ch; while( ch = = 'y' ) { temp = new list; cout<<"\n ENTER DATA TO BE STORED:\n"; cin>>temp->data; temp->next = NULL; curr->next = temp; curr = temp; cout<<"\n DO YOU WANT TO INSERT ANOTHER NODE (Y/N):"; cin>>ch; }

79 void slist :: disp ( ) { if( start = = NULL) cout<<"\n LIST IS EMPTY"; else { cout<<"\n DATA PRESENT IN A LIST IS \n"; temp = start; while( temp -> next ! = NULL) { cout data next "; temp = temp -> next; } cout data next<<"|"; } slist::list *slist :: search( int key) { temp = start; while( temp -> next ! = NULL) { if( temp->data = = key ) return temp; else temp = temp->next; } if( temp->next = = NULL )

80 if( temp->data = = key ) return temp; else return NULL; } void slist:: insert ( ) { int key; cout<<"\n ENTER DATA AFTER WHICH WE CAN INSERT NEW NODE:"; cin>>key; add=search(key); if( add = = NULL ) cout<<"\n NODE IS NOT FOUND"; else { temp = new list; cout<<"\n ENTER INSERTED ELEMENT"; cin>>temp->data; if( add->next = = NULL) { temp->next = NULL; add->next = temp; curr = temp; }

81 else { addr = add->next; add->next = temp; temp->next = addr; } void slist :: del ( ) { int key; cout<<"\n ENTER NODE DATA SHOULD BE DELETE:\n"; cin>>key; add = search ( key ); if( add = = NULL ) cout<<"\n NODE IS NOT FOUND\n"; else if( curr = = add ) { curr = start; while( curr->next ! = NULL) { temp = curr; curr = curr->next; }

82 free ( curr ); curr = temp; curr->next = NULL; } else if( start = = add ) { temp = start; start = start->next; free( temp ); } else { tem = add->next; temp = start; while( temp-> next ! = add) temp = temp->next; temp->next = tem; free( add ); } void main ( ) { slist ob; int key, ch; list *temp; clrscr ( ); cout<<"\n * * * SINGLE LINKED LIST OPERATION * * * \n";

83 cout<<"\n 1. CREATE \n 2. DISPLAY \n 3. INSERT \n 4. DELETE \n 5. SEARCH \n 6.EXIT \n"; cout<<"\n *************************\n"; do { cout<<"\n ENTER YOUR CHOICE \n"; cin>>ch; switch ( ch ) { case 1: ob.create ( ); break; case 2: ob.disp ( ); break; case 3: ob.insert ( ); break; case 4: ob.del ( ); break; case 5: cout<<"\n ENTER THE ELEMENT TO SEARCH"; cin>>key; temp=ob.search(key); if( temp = = NULL) cout<<"\n ELEMENT IS NOT FOUND \n"; else cout<<"\n ELEMENT IS FOUND \n"; break; case 6: exit(0); default: cout<<"\n INVALID CHOICE \n"; } }while( ch ! = 0 ); getch ( ); }

84 OUTPUT:

85 SINGLE CIRCULAR LINKED LIST THEORY: The linked list that we have seen so far is often know as linear lists. The elements of such a linked list can be accessed, first by setting up a pointer pointing to the first node in the list and then traversing the entire list using this pointer. Although a linear linked list is a useful data structure, it has several shortcomings.

86 SOURCE CODE: /* Program to implement single circular linked list */ #include class clist { private: struct list { int data; struct list *next; }*start,*temp,*curr,*add,*tem,*addr; public: void init();void creat(); void display();list *search(int); void insert();void del(); }; void clist::init() { start=temp=curr=NULL; }

87 void clist::creat() { char ch; temp=new list; cout<<"\n ENTER ENTER DATA TO BE STORED ::"; cin>>temp->data; cout<<"\nADDRESS OF STARTING NODE :: "<next=start; start=curr=temp; cout<<"\nDO YOU WANT TO INSERT ANOTHER NODE (y/n) :: "; cin>>ch; while(ch=='y') { temp=new list; cout<<"\n ENTER DATA TO BE STORED :: "; cin>>temp->data; temp->next=start; curr->next=temp; curr=temp; cout<<"\nDO YOU WANT TO INSERT ANOTHER NODE (y/n) :: "; cin>>ch; }

88 void clist::display() { if(start==NULL) cout<<"\nLIST IS EMPTY....."; else cout<<"\nDATA PRESENT IN A LIST IS :: \n"; temp=start; while(temp->next!=start) { cout data next "; temp=temp->next; } cout data next<<"|"; } clist::list *clist::search(int key) { temp=start; while(temp->next!=start) { if(temp->data==key) return temp; else temp=temp->next; }

89 if(temp->next==NULL) { if(temp->data==key) return temp; else return NULL; } return NULL; } void clist::insert() { int key; cout<<"\n ENTER DATA AFTER WHICH WE CAN INSERTED NEW NODE :: "; cin>>key; add=search(key); if(add==NULL) cout<<"\n NODE IS NOT FOUND...."; else { temp=new list; cout<<"\n ENTER INSERTED ELEMENT :: "; cin>>temp->data;

90 if(add->next==start) { temp->next=start; add->next=temp; curr=temp; } else { addr=add->next; add->next=temp; temp->next=addr; } void clist::del() { int key; cout<<"\nEnter node to deleted:"; cin>>key; add=search(key); if(add==NULL) cout<<"\nNode is not found"; else if(curr==add)

91 { curr=start; while(curr->next!=start) { temp=curr; curr=curr->next; } free(curr); curr=temp; curr->next=start; } else if(start==add) { temp=start; start=start->next; free(temp); } else { tem=add->next; temp=start; while(temp->next!=add) temp=temp->next; temp->next=tem; free(add); }

92 void main() { clist ob; int key,ch; clist::list *temp; clrscr(); cout<<"\nCIRCULAR LINKED LIST \n"; cout<<"\n1.Create\n2.Display\n3.Insert\n4.Delete\n5.Search\n6.Exit\n"; do { cout<<"\nEnter your choice"; cin>>ch; switch(ch) { case 1:ob.creat(); break; case 2:ob.display(); break; case 3:ob.insert(); break; case 4:ob.del(); break;

93 case 5:cout<<"\nEnter search element"; cin>>key; temp=ob.search(key); if(temp==NULL) cout<<"\nElement is not found"; else cout<<"\nElement is found"; break; case 6:exit(0); default:cout<<"Invalid choice"; } }while(ch!=6); getch(); }

94 OUTPUT:

95 DOUBLE LINKED LIST AIM: Write a program in C++ to implement DOUBLE LINKED LIST THEORY: A two-way list is a linear collection of data elements, called nodes, where each node N is divided into three parts: An item data field. A pointer field next which contains the location of the next node in the list. A pointer field prev which contains the location of the previous node in the list. The list requires two list pointer variables: FIRST, which points to the first node in the list, and LAST, which points to the last node in the list. The figure contains a schematic diagram of such a list. Observe that the null pointer appears in the next field of the last node in the list and also in the prev field of the first node in the list. Observe that, using the variable FIRST and the pointer field next, we can traverse a two-way list in the forward direction as before. On the other hand, using the variable LAST and the pointer field prev, we can also traverse the list in the backward direction.

96 OPERATION ON TWO-WAY LISTS: 1. Traversing. 2. Searching. 3. Deleting 4Inserting

97 SOURCE CODE: /* Program to implement Double linked list */ #include class dlist { private: struct list { int data; struct list *next,*prev; }*start,*temp,*curr,*add,*addr,*tem; public: void init(); void creat();void display(); list *search(int); void insert();void del(); }; void dlist::init() { start=temp=curr=NULL; }

98 void dlist::creat() { char ch; temp=new list; cout<<"\nENTER DATA TO BE STORED :: "; cin>>temp->data; cout<<"\n STARTING NODE ADDRESS :: "<next=NULL; temp->prev=NULL; start=curr=temp; cout<<"\n DO YOU WANT TO INSERT ANOTHER NODE (y/n) :: "; cin>>ch; while(ch=='y') { temp=new list; cout<<"\nENTER DATA TO BE STORED :: "; cin>>temp->data; temp->next=NULL; temp->prev=curr; curr->next=temp; curr=temp; cout<<"\nDO YOU WANT TO INSERT ANOTHER NODE (y/n) :: "; cin>>ch; }

99 void dlist::display() { if(start==NULL) cout<<"\n LIST IS EMPTY...."; else { cout<<"\n DATA PRESENT IN A LIST\n:::"; temp=start; while(temp->next!=NULL) { cout prev data next "; temp=temp->next; } cout prev data next<<"|"; } dlist::list *dlist::search(int key) { temp=start; while(temp->next!=NULL) { if(temp->data==key)

100 return temp; else temp=temp->next; } if(temp->next==NULL) { if(temp->data==key) return temp; else return NULL; } return NULL; } void dlist::insert() { int key; cout<<"\nENTER DATA AFTER WHICH WE CAN INSERT A NEW NODE :: "; cin>>key; add=search(key); if(add==NULL) cout<<"\n NODE IS NOT FOUND....."; else tem=new list;

101 cout<<"\n ENTER ELEMENT TO BE SEARCHED :: "; cin>>tem->data; if(add->next==NULL) { tem->next=NULL; tem->prev=add; add->next=tem; curr=tem; } else { addr=add->next; add->next=tem; tem->next=addr; tem->prev=add; } void dlist::del() { int key; cout<<"\n ENTER NODE DATA TO BE DELETED :: "; cin>>key; add=search(key); if(add==NULL) cout<<"\n NODE IS NOT FOUND :: "; else

102 if(curr==add) { curr=start; while(curr->next!=NULL) { temp=curr; curr=curr->next; } free(curr); curr=temp; curr->next=NULL; } else if(start==NULL) { temp=start; start=start->next; free(temp); } else { tem=add->next; temp=start; while(temp->next!=add) temp=temp->next; temp->next=tem; free(add); }

103 void main() { dlist ob; int key,ch; dlist::list *temp; clrscr(); cout<<"**********DOUBLE LINKED LIST**********"; cout<<"\n1.Create\n2.Display\n3.Insert\n4.Delete\n5.Search\n6.Exit\n"; do { cout<<"\nENTER YOUR CHOICE :: "; cin>>ch; switch(ch) { case 1:ob.creat(); break; case 2:ob.display(); break; case 3:ob.insert(); break; case 4:ob.del(); break;

104 case 5:cout<<"\n ENTER SEARCH ELEMENT :: "; cin>>key; temp=ob.search(key); if(temp==NULL) cout<<"\n ELEMENT IS NOT FOUND...."; else cout<<"\n ELEMENT IS FOUND....."; break; case 6:exit(0); default:cout<<"\n INVALID CHOICE...."; } }while(ch!=6); getch(); }

105 OUTPUT:

106 GRAPH TRAVERSING: DEPTH FIRST SEARCH THEORY: DFS is an uninformed search that progresses by expanding the first child node of the search tree that appears and thus going deeper and deeper until a goal node is found, or until it hits a node that has no children. Then the search backtracks, returning to the most recent node it hadn't finished exploring. In a non-recursive implementation, all freshly expanded nodes are added to a LIFO stack for exploration. Space complexity of DFS is much lower than BFS (breadth-first search). It also lends itself much better to heuristic methods of choosing a likely-looking branch. Time complexity of both algorithms are proportional to the number of vertices plus the number of edges in the graphs they traverse (O(|V| + |E|)).

107 SOURCE CODE: /*Program To Implement Depth First Search */ #include #define MAX 20 class depth { private: int a[MAX][MAX], visited[MAX]; int n, top; public: void init ( ); void input ( ); void dfs ( int ); }; void depth::init ( ) { int i, j; for( i = 0; i < MAX; i + + ) { visited[ i ] = 0; for( j =0; j < MAX ; j + + ) a[ i ] [ j ] = 0; } top = - 1; } void depth::input ( ) { int i, j; cout<<"\nENTER NUMBER OF NODES IN A GRAPH :: ";

108 cin>>n; cout<<"\nENTER ADJACENCY MATRIX FOR A GRAPH :: \n"; for( i = 1; i <= n; i + +) for( j = 1; j <= n; j + + ) cin>>a[ i ][ j ]; } void depth::dfs ( int v) { int i; visited[v] = 1; cout "; for( i = 1; i <= n; iI + + ) if( a [ v ] [ i ] = = 1 && visited [ i ] = = 0) dfs ( i ); } void main ( ) { depth ob; int start; clrscr ( ); ob.init ( ); ob.input ( ); cout<<"\nSTARTING NODE FOR DFS TRAVERSING :: "; cin>>start; cout<<"\nDEPTH FIRST SEARCH TRAVERSING IS ::\n\n"; ob.dfs ( start ); getch ( ); }

109 OUTPUT:

110 GRAPH TRAVERSING: BREADTH FIRST THEORY: BFS is an uninformed search method that aims to expand and examine all nodes of a graph or combinations of sequence by systematically searching through every solution. In other words, it exhaustively searches the entire graph or sequence without considering the goal until it finds it. From the standpoint of the algorithm, all child nodes obtained by expanding a node are added to a FIFO queue. In typical implementations, nodes that have not yet been examined for their neighbors are placed in some container (such as a queue or linked list) called "open" and then once examined are placed in the container "closed".  Algorithm for Breadth First Search 1.Enqueue the root node. 2.Dequeue a node and examine it. If the element sought is found in this node, quit the search and return a result. Otherwise enqueue any successors (the direct child nodes) that have not yet been discovered. 3.If the queue is empty, every node on the graph has been examined – quit the search and return "not found". 4.Repeat from Step 2.

111 SOURCE CODE: /*Program To Implement Breadth First Search */ #include #define MAX 20 class breadth { private: int a[MAX][MAX], visited[MAX], queue[50]; int n, front, rear; public: void init ( ); void input ( ); void bfs ( ); }; void breadth::init ( ) { int i, j; for( i = 0; i < MAX; i + + ) { visited [ i ] = 0; for( j = 0; j < MAX; j + + ) a[ i ] [ j ] = 0; } front = rear = - 1; }

112 void breadth::input ( ) { int i, j; cout<<"\nENTER NUMBER OF NODES IN A GRAPH :: "; cin>>n; cout<<"\nENTER ADJACENCY MATRIX FOR A GRAPH :: \n"; for( i = 1;i <= n; i + + ) for( j = 1; j <= n; j + + ) cin>>a[ i ][ j ]; } void breadth::bfs ( ) { int i, start; cout<<"\nSTARTING NODE FOR BFS TRAVERSING :: "; cin>>start; cout<<"\n BREADTH FIRST SEARCH TRAVERSING IS:: \n \t"; cout<

113 for( i =1; i <= n; i + + ) { if(a[ start ][ i ] = =1 && visited[ i ] = = 0) { cout "<

114 OUTPUT:

115 SHORTEST PATH FOR GRAPH THEORY: Shortest path is nothing but the path which lies between two nodes with the lowest cost. In a graph contain so many paths are existed between two nodes (source node to destination node) to choose the lowest cost path to reach from source node to destination node is nothing but shortest path algorithm. Shortest path algorithm was first proposed by E. W. DIJKSTRA. SOURCE CODE: /*Program To Implement Shortest Path for Graph */ #include #define INF 9999 class stpath { private: int i, j, k; public: void spath(int [ ][20], int ); void display(int [ ][20], int ); };

116 void stpath::spath(int a[ ][20], int n) { for( i = 0 ;i < n; I + + ) for( j = 0; j < n; j + + ) if(a[ i ] [ j ] = = 0) a[ i ][ j ] = INF; cout<<"\nADJACENCY MATRIX OF COST OF EDGES ARE ::"; display( a, n ); for( k = 0; k < n; k + + ) for( i = 0; i < n; i + + ) for( j = 0; j < n; j + + ) if( a[ i ][ j ] > a[ i ] [ k] + a[ k ][ i ]) a[ i ][ j ] = a[ i ][ k ] + a[ k ][ j ]; cout<<"\nADJACENCY MATRIX OF LOWEST COST OF EDGES ARE ::\n"; display(a, n); } void stpath::display(int a[ ] [20], int n) { for( i = 0; i < n; i + + ) { for( j = 0; j < n; j + + ) cout<

117 void main() { int i, j, n, a[20][20]; stpath ob; clrscr(); cout<<"\nENTER NUMBER OF NODES IN A GRAPH :: "; cin>>n; cout<<"\nENTER ADJACENCY MATRIX ::\n"; for( i = 0; i < n; i + + ) for( j = 0; j < n; j + + ) { cout<<"Enter "<>a[ i ] [ j ]; } ob.spath(a, n); getch ( ); }

118 OUTPUT:


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