Computer Project For ISC

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COMPUTER SCEIENCE

COMPUTER

SCEIENCE

PROJECT

NAME : TABISH HAIDER RIZVI

CLASS : XII B

ROLL NO : 25

ACKNOWLEDGEMENT

I would like to thank my Computer Science

Teacher ,

Mr. J.V. Nagendra Rao, who guided me in making this

project by giving

some valuable points.

I would also like to thank my parents who

helped me in

TABLE OF CONTENTS

S.No. PROGRAM PAGE No. 1 TO create Pascal’s triangle 4

2 To display entered number in words 5 3 To display A.P. series and its sum 6 4 To display calendar of any month of any year 8 5 To calculate factorial using recursion 10 6 To display Fibonacci series using recursion 11 7 To calculate GCD using recursion 12 8 To display spiral matrix 13 9 To display magical square 15 10 To search an array using Linear Search 17 11 To search an array using Binary Search 20 12 To sort an array using Selection sort 22 13 To sort an array using Bubble sort 24 14 To convert a decimal no into it binary equivalent 27 15 To display date from entered day no. 28 16 To create a pattern from entered string 30 17 To check if entered string is palindrome or not 31 18 To display a frequency of each character in entered string 32 19 To find a word in entered string 34 20 To decode the entered string 36 21 To display the entered string in alphabetical order. 38 22 To create a string and count number of vowels and consonants. 40 23 To create a string and count number of words 41 24 To create a string and replace all vowels with * 42 25 To create a double-dimensional array of 4*4 subscripts. 43 26 To generate sum of all elements of a double dimensional array of 5*5 subscripts 44 27

To generate product of two arrays of 5 subscripts as a third array

46

28

To find sum of each column of a double dimensional array

47

29 To find sum of diagonal of a double dimensional array of 4*4 subscripts 49 30 To calculate the commission of a salesman 51

PROGRAM 1 - T

create Pascal’s triangle

Algorithm:

STEP 1 - START STEP 2 - pas[0] = 1

STEP 3 - IF i=0 THEN GOTO STEP 4 STEP 4 - IF j=0 THEN GOTO STEP 5 STEP 5 - PRINT pas[j]+" "

STEP 6 - i++& IF i<n GOTO STEP 4 STEP 7 - j=0 & IF j<=i GOTO STEP 5 STEP 8 - IF j=i+1 THEN GOTO STEP 7 STEP 9 - pas[j]=pas[j]+pas[j-1]

STEP 10 - j--& IF j>0 GOTO STEP 9 STEP 11 - END

Solution:

class pascal {

public void pascalw(int n) {

int [ ] pas = new int [n+1]; pas[0] = 1;

for (int i=0; i<n; i++) {

for (int j=0; j<=i; ++j)

System.out.print(pas[j]+" "); System.out.println( );

for (int j=i+1; j>0; j--) pas[j]=pas[j]+pas[j-1]; }

}

Output:

PROGRAM 2 -

To display entered number in words

Algorithm:

STEP 1 - START STEP 2 - INPUT amt

STEP 3 - z=amt%10 , g=amt/10

STEP 4 - IF g!=1 THEN GOTO STEP 5 OTHERWISE GOTO STEP 6 STEP 5 - PRINT x2[g-1]+" "+x1[z]

STEP 6 - PRINT x[amt-9 STEP 7 - END

SOLUTION:

import java.io.*; class eng

{

public static void main(String args[])throws IOException {

BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

String x3;

int amt=Integer.parseInt(br.readLine()); int a,b,c,y,z,g; String x[]={" “,"Ten","Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen", "Seventeen","Eighteen","Nineteen"}; String x1[]={" ","One","Two","Three","Four","Five","Six","Seven","Eight","Nine"}; String x2[]={" ","Twenty","Thirty","Fourty","Fifty","Sixty","Seventy","Eighty","Ninety"}; z=amt%10; g=amt/10; if(g!=1) System.out.println(x2[g-1]+" "+x1[z]); else System.out.println(x[amt-9]); } }

Output:

Enter any Number(less than 99) 45

Fourty Five

PROGRAM 3

-

To display A.P. series and its sum

Algorithm :

STEP 1 - START STEP 2 - a = d = 0 STEP 3 - IMPORT a, d

STEP 4 - this.a = a & this.d = d STEP 5 - IMPORT n

STEP 6 - RETURN (a+(n-1)*d) STEP 7 - IMPORT n

STEP 8 - RETURN (n*(a+nTHTerm(n))/2) STEP 9 - IMPORT n

STEP 10 - PRINT \n\tSeries\n\t"

STEP 12 - PRINT nTHTerm(i)+" "

STEP 13 - i++ & IF i<=n GOTO STEP 12 STEP 14 - PRINT n\tSum : "+Sum(n) STEP 15 - END

SOLUTION:

class APSeries {

private double a,d; APSeries() { a = d = 0; } APSeries(double a,double d) { this.a = a; this.d = d; } double nTHTerm(int n) { return (a+(n-1)*d); } double Sum(int n) { return (n*(a+nTHTerm(n))/2); } void showSeries(int n) { System.out.print("\n\tSeries\n\t"); for(int i=1;i<=n;i++) { System.out.print(nTHTerm(i)+" "); } System.out.print("\n\tSum : "+Sum(n)); } }

Output:

a = 5 d=2 n=10 Series 5.0 7.0 9.0 11.0 13.0 15.0 17.0 19.0 21.0 23.0 Sum : 140.0

PROGRAM 4

-

To display calendar of any month of

any year

Algorithm :

STEP 1 - START

STEP 2 - INPUT int month,int year

STEP 3 - int i,count=0,b,c,d=1 & String w="SMTWTFS"

STEP 4 - IF (year%100==0 && year%400==0) || (year%100!=0 && year %4==0)

STEP 5 - days[1]=29

STEP 6 - PRINT "================The Calendar of "+month1[month-1]+" "+year+" is==================") STEP 7 - IF i=0 THEN GOTO STEP 8

STEP 8 - PRINT (i)+"\t" & " " STEP 9 - IF i=1 GOTO STEP 10

STEP 10 - IF (year%100==0 && year%400==0) || (year%100!=0 && year %4==0)THEN GOTO STEP 11OTHERWISE GOTO STEP 12

STEP 11 - count+=2 STEP 12 - count+=1

STEP 13 - IF i=0 GOTO STEP 14

STEP 14 - count+=days[i] , count+=1, count%=7 & b=7-count STEP 15 - IF b!=1 || b!=7 GOTO STEP 16

STEP 16 - IF count>0 GOTO STEP 17,18 STEP 17 - PRINT ' '+"\t")

STEP 18 -

count--STEP 19 - IF i=1 GOTO count--STEP 20

STEP 20 - IF b>0 && IF d<=days[month-1] GOTO STEP 21,22 STEP 21 - PRINT d+"\t"

STEP 22 - d++ & b--STEP 23 - b=7

STEP 24 - i++ & IF i<MONTH GOTO STEP14 STEP 25 - PRINT " "

SOLUTION:

class calendar {

public void dee(int month,int year) { int i,count=0,b,c,d=1; String w="SMTWTFS"; int days[]={31,28,31,30,31,30,31,31,30,31,30,31}; String month1[]={"January","February","March","April","May","June","July", "August","September","October","November","December"} ;

If((year%100==0 && year%400==0) || (year%100!=0 && year%4==0)) days[1]=29; System.out.println("================The Calendar of "+month1[month-1]+" "+year+" is=================="); for(i=0;i<w.length();i++) System.out.print(w.charAt(i)+"\t"); System.out.println(" "); for(i=1;i<year;i++)

if((year%100==0 && year%400==0) || (year%100!=0 && year %4==0)) count+=2; else count+=1; for(i=0;i<month;i++) count+=days[i]; count+=1; count%=7; b=7-count; if(b!=1 || b!=7) while(count>0) { System.out.print(' '+"\t"); count--; } for(i=1;i<7;i++) {

while(b>0 && d<=days[month-1]) {

System.out.print(d+"\t"); d++;

} b=7; System.out.println(" "); } } }

Output:

=====The Calendar of October 2009 is====

S M T W T F S 1 2 3 4 5 6 7 8 9 1 0 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

PROGRAM 5

– To calculate factorial using recursion

Algorithm :

STEP 1 - START STEP 2 - INPUT n

STEP 3 - IF(n<2) THEN return 1 OTHERWISE return (n * fact(n-1)) STEP11 - END

SOLUTION:

import java.io.*; class factorial {

public static void main(String args[]) throws IOException {

InputStreamReader(System.in));

System.out.println("enter no ="); int n = Integer.parseInt(br.readLine()); factorial obj = new factorial();

long f = obj.fact(n);

System.out.println("factotial ="+f); }

public long fact(int n) { if(n<2) return 1; else return (n*fact(n-1)); } }

Output:

PROGRAM 6

-

To display Fibonacci series using

recursion

Algorithm :

STEP 1 - START STEP 2 - INPUT n

STEP 3 - IF(n<=1) THEN return 1 OTHERWISE return (fib(n-1) +fib(n-2)) STEP11 - END

SOLUTION:

import java.io.*; class fibonacci

{

public static void main(String args[]) throws IOException {

fibonacci obj = new fibonacci();

BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); System.out.println("enter no of term ="); int n = Integer.parseInt(br.readLine()); System.out.println(); for(int i=1;i<=n;i++) { int f = obj.fib(i); System.out.print(f+" "); } }

public int fib(int n) {

if(n<=1) return n; else

return (fib(n-1) +fib(n-2)); } }

Output:

PROGRAM 7

-

T

o calculate GCD using recursion

Algorithm :

STEP 1 - START STEP 2 - INPUT p,q

STEP 3 - IF(q=0) THEN return p OTHERWISE return calc(q,p%q) STEP11 - END

SOLUTION:

public static void main(String args[]) throws IOException {

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("enter the numbers ="); int p = Integer.parseInt(br.readLine()); int q = Integer.parseInt(br.readLine()); gcd obj = new gcd(); int g = obj.calc(p,q); System.out.println("GCD ="+g); }

public int calc(int p,int q) { if(q==0) return p; else return calc(q,p%q); } }

Output :

enter the numbers = 12

8

PROGRAM 8

-

To display spiral matrix.

Algorithm:

STEP 1 - START STEP 2 - INPUT a[][]

STEP 3 - IF p!=(int)Math.pow(l,2) GOTO STEP 4 STEP 4 - IF co!=0 GOTO STEP 5

STEP 5 - re=1

STEP 6 - IF ri=1;ri<=k1-re;ri++ GOTO STEP 7 STEP 7 - p++,c++

STEP 8 - IF c==l GOTO STEP 9 STEP 9 - BREAK

STEP 10 - a[r][c]=p

STEP 11 - IF c==l GOTO STEP 12 STEP 12 - BREAK

STEP 13 - IF dw=1 GOTO STEP 14 STEP 14 - p++,r++,a[r][c]=p STEP 15 - IF le=1 GOTO STEP 16 STEP 16 - p++,c--,a[r][c]=p

STEP 17 - IF up=1 GOTO STEP 18 STEP 18 - p++,r--,a[r][c]=p

STEP 19 - k1=k1+2, k2=k2+2 & co++

STEP 20 - up++ & IF up<=k2-1 GOTO STEP 18 STEP 21 - le++ & IF le<=k2-1 GOTO STEP 16 STEP 22 - dw++ & IF dw<=k1-1 GOTO STEP 14 STEP 23 - IF y=0 GOTO STEP 24

STEP 24 - IF yy=0 GOTO STEP 25 STEP 25 - PRINT "\t"+a[y][yy]) & ()

STEP 26 - yy++ & IF yy<l GOTO STEP 25 STEP 27 - y++ & IF y<l GOTO STEP 24 STEP 28 - END

SOLUTION:

import java.io.*; class spiralsm {

public static void main(String[] args) throws IOException {

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("enter the dimension of matrix ="); int l = Integer.parseInt(br.readLine());

a=new int[l][l]; r=l/2;c=r-1; if(l%2==0) {

System.out.println("wrong entry for spiral path"); System.exit(0); } while(p!=(int)Math.pow(l,2)) { if(co!=0) re=1; for(int ri=1;ri<=k1-re;ri++) {p++;c++;if(c==l)break;a[r][c]=p;} if(c==l)break; for(int dw=1;dw<=k1-1;dw++) {p++;r++;a[r][c]=p;} for(int le=1;le<=k2-1;le++) {p++;c--;a[r][c]=p;} for(int up=1;up<=k2-1;up++) {p++;r--;a[r][c]=p;} k1=k1+2; k2=k2+2; co++; } for(int y=0;y<l;y++) { for(int yy=0;yy<l;yy++) System.out.print("\t"+a[y][yy]); System.out.println(); System.out.println(); } } }

Output:

5 21 22 23 24 25 20 7 8 9 10 19 6 1 2 11 18 5 4 3 12 17 16 15 14 13

PROGRAM 9

-

T

o display magical square

Algorithm:

STEP 1 - START

STEP 2 - arr[][]=new int[n][n],c=n/2-1,r=1,num

STEP 3 - IF num=1;num<=n*n;num++ GOTO STEP 4 STEP 4 - r--,c++

STEP 5 - IF r==-1 GOTO STEP 6 STEP 6 - r=n-1

STEP 7 - IF c>n-1 GOTO STEP 8 STEP 8 - c=0

STEP 9 - IF arr[r][c]!=0 GOTO STEP 10 STEP 10 - r=r+2 &

c--STEP 11 - num++ & IF num<=n*n GOTO c--STEP 4 STEP 12 - arr[r][c]=num

STEP 13 - IF r==0&&c==0 GOTO STEP 14 STEP 14 - r=n-1, c=1 & arr[r][c]=++num STEP 15 - IF c==n-1&&r==0 GOTO STEP 16 STEP 16 - arr[++r][c]=++num

STEP 17 - PRINT ()

STEP 18 - IFr=0 GOTO STEP 19 STEP 19 - IF c=0 GOT STEP 20 STEP 20 - PRINT arr[r][c]+" " & () STEP 21 - c++ & IF c<n GOTO STEP 20 STEP 21 - r++ & r<n GOTO STEP 19 STEP 22 - END

import java.io.*; class magicalsquare {

public static void main(String args[])throws Exception {

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("enter the dimension of magical square="); int n = Integer.parseInt(br.readLine());

int arr[][]=new int[n][n],c=n/2-1,r=1,num; for(num=1;num<=n*n;num++) { r--; c++; if(r==-1) r=n-1; if(c>n-1) c=0; if(arr[r][c]!=0) { r=r+2; c--; } arr[r][c]=num; if(r==0&&c==0) { r=n-1; c=1; arr[r][c]=++num; } if(c==n-1&&r==0) arr[++r][c]=++num; } System.out.println(); for(r=0;r<n;r++) { for(c=0;c<n;c++) System.out.print(arr[r][c]+" "); System.out.println(); } } }

Output :

enter the dimension of magical square= 5 17 24 1 8 15 23 5 7 14 16 4 6 13 20 22 10 12 19 21 3 11 18 25 2 9

PROGRAM 10

-

T

o search an array using Linear

Search

Algorithm:

STEP 1 - START STEP 2 - INPUT a[]

STEP 3 - FROM i=0 to i<n REPEAT STEP 4 STEP 4 - PRINT a[i]+" "

STEP 5 - flag=-1

STEP 6 - FROM i=0 to i<n REPEAT STEP 7 STEP 7 - IF (a[i] == v) THEN flag =i

STEP 8 - IF (flag=-1) THEN GOTO STEP 9 OTHERWISE GOTO STEP 10 STEP 9 - PRINT “ not found”

STEP 10 - PRINT v+" found at position - "+flag STEP 11 - END

SOLUTION:

int a[] = new int[100];

static BufferedReader br =new BufferedReader(new InputStreamReader(System.in));

public linear_search(int nn) {

n=nn; }

public void input() throws IOException { System.out.println("enter elements"); for(i=0;i<n;i++) { a[i] = Integer.parseInt(br.readLine()); } }

public void display() { System.out.println(); for(i=0;i<n;i++) { System.out.print(a[i]+" "); } }

public void search(int v) {

int flag=-1;

for(int i=0; i<n ; i++) { if(a[i] == v) flag =i; } if(flag== -1 ) System.out.println("not found"); else

System.out.println(v+" found at position - "+flag);

}

public static void main(String args[]) throws IOException {

linear_search obj = new linear_search(10); obj.input();

obj.display();

System.out.println("enter no. to be searched -"); int v = Integer.parseInt(br.readLine());

} }

Output :

enter no. to be searched -1

1 found at position - 4

PROGRAM 11

-

T

o search an array using Binary

Search

Algorithm:

STEP 1 - START STEP 2 - INPUT a[]

STEP 3 - FROM i=0 to i<n REPEAT STEP 4 STEP 4 - PRINT a[i]+" "

STEP 5 - flag=-1 , l=0, u=n-1

STEP 6 - IF(l<=u && flag=-1) REPEAT STEP 7 AND Step 8 STEP 7 - m = (l+u)/2

STEP 8 - IF (a[m] == v) THEN flag =m OTHERWISE GOTO STEP 9 STEP 9 - IF (a[m] < v) THEN l = m+1 OTHERWISE u =m-1

STEP 10 - IF (flag=-1) THEN GOTO STEP 11 OTHERWISE GOTO STEP 12 STEP 11 - PRINT “ not found”

STEP 12 - PRINT v+" found at position - "+flag STEP 13 - END

STEP 11 - END

SOLUTION:

import java.io.*; class binary_search {

int n,i;

int a[] = new int[100];

static BufferedReader br =new BufferedReader(new InputStreamReader(System.in)); public binary_search(int nn) { n=nn; }

public void input() throws IOException { System.out.println("enter elements"); for(i=0;i<n;i++) { a[i] = Integer.parseInt(br.readLine()); } }

public void display() { System.out.println(); for(i=0;i<n;i++) { System.out.print(a[i]+" "); } }

public void search(int v) {

int l=0; int u = n-1; int m; int flag=-1;

while( l<=u && flag == -1) { m = (l+u)/2; if(a[m] == v) flag = m;

else if(a[m] < v) l = m+1; else u = m-1; } if(flag== -1 ) System.out.println("not found"); else

System.out.println(v+" found at position - "+flag);

}

public static void main(String args[]) throws IOException {

binary_search obj = new binary_search(10); obj.input();

obj.display();

System.out.println("enter no. to be searched -"); int v = Integer.parseInt(br.readLine()); obj.search(v); } }

Output :

-1

1 found at position - 4

PROGRAM 12

-

T

o sort an array using Selection

sort

Algorithm:

STEP 1 - START STEP 2 - INPUT a[]

STEP 3 - FROM i=0 to i<n REPEAT STEP 4 STEP 4 - PRINT a[i]+" "

STEP 5 - flag=-1

STEP 6 - FROM i=0 to i<n-1 REPEAT STEP 7 to STEP 11 STEP 7 - min =i

STEP 8 - FROM j=i+1 to j<n REPEAT STEP 8 STEP 9 - IF(a[j]<a[min]) then min =j STEP 10 - IF (min!=i) GOTO STEP 11

STEP 11 - temp = a[i], a[i] =a[min], a[min] = temp

SOLUTION:

import java.io.*; class selection_sort {

int n,i;

int a[] = new int[100]; public selection_sort(int nn) { n=nn; }

public void input() throws IOException {

BufferedReader br =new BufferedReader(new InputStreamReader(System.in)); System.out.println("enter elements"); for(i=0;i<n;i++) { a[i] = Integer.parseInt(br.readLine()); }

}

public void display() { System.out.println(); for(i=0;i<n;i++) { System.out.print(a[i]+" "); } }

public void sort() { int j,temp,min; for(i=0;i<n-1;i++) { min =i; for(j=i+1;j<n;j++) { if(a[j]<a[min]) min =j; } if(min!=i) { temp = a[i]; a[i] =a[min]; a[min] = temp; } } }

public static void main(String args[]) throws IOException {

selection_sort x = new selection_sort(5); x.input(); System.out.print("Before sorting - "); x.display(); System.out.print("After sorting - "); x.sort(); x.display(); }

}

Output:

PROGRAM 13

-

T

o sort an array using Bubble Sort

Algorithm:

STEP 1 - START STEP 2 - INPUT a[]

STEP 3 - FROM i=0 to i<n REPEAT STEP 4 STEP 4 - PRINT a[i]+" "

STEP 5 - flag=-1

STEP 6 - FROM i=0 to i<n-1 REPEAT STEP 7 to STEP 9 STEP 7 - FROM j=i+1 to j<n REPEAT STEP 8

STEP 8 - IF(a[j] > a[j+1]) THEN GOTO STEP 9 STEP 9 - temp = a[i], a[i] =a[min], a[min] = temp STEP 10 - END

SOLUTION:

int a[] = new int[100];

public bubble_sort(int nn) {

}

public void input() throws IOException {

BufferedReader br =new BufferedReader(new InputStreamReader(System.in)); System.out.println("enter elements"); for(i=0;i<n;i++) { a[i] = Integer.parseInt(br.readLine()); } }

public void display() { System.out.println(); for(i=0;i<n;i++) { System.out.print(a[i]+" "); } }

public void sort() {

int j,temp;

for(i=0 ; i<n-1 ; i++) { for(j=0 ; j<n-1-i ; j++) { if(a[j] > a[j+1]) { temp = a[j]; a[j] =a[j+1]; a[j+1] = temp; } } } }

public static void main(String args[]) throws IOException {

bubble_sort x = new bubble_sort(5); x.input(); System.out.print("Before sorting - "); x.display(); System.out.print("After sorting - "); x.sort(); x.display(); } }

Output:

PROGRAM 14

-

To convert a decimal no into it

binary equivalent

SOLUTION:

STEP 1 - START STEP 2 - n = 30 STEP 3 - INPUT int no STEP 4 - c =0 , temp = no

STEP 5 - IF (temp!=0) REPEAT STEP 6

STEP 6 - a[c++] = temp%2, temp = temp / 2 STEP 7 - FROM i=c-1 to i>0 REPEAT STEP 8 STEP 8 - PRINT a[i]

STEP 9 - END Program :

import java.io.*; class dec_bin {

int n,i;

int a[] = new int[100];

static BufferedReader br =new BufferedReader(new InputStreamReader(System.in)); public dec_bin(int nn) { n=nn; }

public void dectobin(int no) {

int c = 0; int temp = no; while(temp != 0) { a[c++] = temp % 2; temp = temp / 2; }

System.out.println("Binary eq. of "+no+" = "); for( i = c-1 ; i>=0 ; i--)

System.out.print( a[ i ] ); }

public static void main(String args[]) throws IOException {

dec_bin obj = new dec_bin(30);

System.out.println("enter decimal no -"); int no = Integer.parseInt(br.readLine()); obj.dectobin(no); } }

Output :

Binary eq. of 56 = 111000

PROGRAM

15

-

To display

date from entered day

no.

SOLUTION:

STEP 1 - START

STEP 2 - INITIALISE a[ ] , m[ ] STEP 3 - INPUT n , yr

STEP 4 - IF ( yr%4=0) THEN a[1] = 29 STEP 5 - t =0 , s = 0

STEP 6 - IF ( t<n) REPEAT STEP 7 STEP 7 - t =t + a[s++]

STEP 8 - d = n + a[--s] - t

STEP 9 - IF ( d ==1|| d == 21 || d == 31 ) then PRINT d + "st" + m[s] + " , "+yr

STEP 10 - IF ( d ==2|| d == 22 ) then PRINT d + "nd" + m[s] + " , "+yr STEP 11 - IF ( d ==3|| d == 23 ) then PRINT d + "rd" + m[s] + " , "+yr

OTHERWISE GOTO STEP 12 STEP 12 - PRINT d + "th" + m[s] + " , "+yr STEP 13 - END

import java.io.*; class daytodate {

static BufferedReader br =new BufferedReader(new InputStreamReader(System.in));

public void calc(int n, int yr) {

int a[ ] = { 31,28,31,30,31,30,31,31,30,31,30,31 } ;

String m[ ] = { "Jan", "Feb", "Mar","Apr","May","Jun","Jul","Aug", "Sep","Oct","Nov","Dec" } ;

if ( yr % 4 == 0) a[1] =29;

int t=0,s=0; while( t < n) { t =t + a[s++]; } int d = n + a[--s] - t; if( d == 1|| d == 21 || d == 31 ) { System.out.println( d + "st" + m[s] + " , "+yr); } if( d == 2 || d == 22 ) { System.out.println( d + "nd" + m[s] + " , "+yr); } if( d == 3|| d == 23 ) { System.out.println( d + "rd" + m[s] + " , "+yr); } else { System.out.println( d + "th" + m[s] + " , "+yr); } }

public static void main(String args[]) throws IOException {

daytodate obj = new daytodate();

System.out.println( "Enter day no = "); int n = Integer.parseInt(br.readLine()); System.out.println( "Enter year = "); int yr = Integer.parseInt(br.readLine()); obj.calc(n,yr); } }

Output:

192

Enter year = 2009

11th Jul , 2009

PROGRAM

16

– To create a pattern from e

ntered

string

Program :

import java.io.*; class pattern {

public static void main (String args[]) throws IOException {

int i,sp,j,k,l;

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("enter the string ="); String s = br.readLine(); l=s.length(); for(i=0;i<l;i++) if(i==l/2) System.out.println(s); else { sp=Math.abs((l/2)-i); for(j=sp;j<l/2;j++) System.out.print(" "); k=0; while(k<3) { System.out.print(s.charAt(i)); for(j=0;j<sp-1;j++) System.out.print(" "); k++; } System.out.println(" "); } } }

Output:

enter the string = india i i i nnn india iii a a a

PROGRAM

17

-

To check if entered string is

palindrome or not

Algorithm :

STEP 1 - START

STEP 2 - INPUT string s STEP 3 - StringBuffer sb = s STEP 4 - sb.reverse

STEP 5 - String rev = sb

STEP 6 - IF rev = s GOTO STEP 7 OTHERWISE GOTO STEP 8 STEP 7 - PRINT " Palindrome"

STEP 8 - PRINT " Not Palindrome" STEP 9 - END

Solution:

import java.io.*; class palindrome {

public static void main(String args[]) throws IOException {

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("enter the string="); String s = br.readLine();

StringBuffer sb = new StringBuffer(s); sb.reverse();

if(s.equals(rev)) System.out.println("Palindrome " ); else System.out.println("Not Palindrome " ); } }

Output :

enter the string= arora

Palindrome

PROGRAM 18 -

To display

a frequency of each

character in entered string

Algorithm :

STEP 1 - START STEP 2 - INPUT str STEP 3 - l=str.length() STEP 4 - PRINT str

STEP 5 - IF i=0 THEN GOTO STEP 4 OTHERWISE GOTO STEP 22 STEP 6 - char a=str.charAt(i)

STEP 7 - IF ii=0 THEN GOTO STEP 4 OTHERWISE GOTO STEP 22 STEP 8 - char b = str.charAt(ii)

STEP 9 - IF a==b GOTO STEP 10 STEP 10 - freq=freq+1

STEP 11 - ii++ & IF ii<1 GOTO STEP 8 STEP 12 - i++ & IF i<1 GOTO STEP 6

STEP 13 - DISPLAY a+" occurs "+freq+" times" STEP 14 - END

Solution :

import java.io.*; class frequency {

private int i,a1,l,p,j,freq;

public frequency() {

p=0;

freq=0;// initialise instance variables

}

public void count(String str) { int ii; l=str.length(); System.out.print(str); for(i=0;i<l;i++) { char a=str.charAt(i); for(ii=0;ii<l;ii++) { char b = str.charAt(ii); if (a==b) freq=freq+1; }

System.out.println(a+" occurs "+freq+" times"); freq=0;

} }

public static void main(String args[]) throws IOException {

BufferedReader br =new BufferedReader(new InputStreamReader(System.in));

System.out.println("enter string"); String str = br.readLine();

frequency x = new frequency(); x.count(str);

} }

Output:

PROGRAM

19

-

To find a word in entered string

Algorithm

STEP 1 - START

STEP 2 - INPUT string s

STEP 3 - StringTokenizer st = s STEP 4 - l =str.length()

STEP 5 - INPUT look STEP 6 - flag = -1

STEP 7 - IF (st.hasMoreElements()) REPEAT STEP 8

STEP 8 - IF (look.equals(st.nextElement())) THEN flag =1 STEP 9 - IF flag = - 1 GOTO STEP 10 OTHERWISE STEP 11 STEP 10 - PRINT "word not found"

STEP 11 - PRINT "word found" STEP 12 - END

Solution:

import java.util.StringTokenizer; import java.io.*;

public class word_search {

public static void main(String[] args) throws IOException {

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("enter the string="); String s = br.readLine();

StringTokenizer st = new StringTokenizer(s," ");

System.out.println("enter the word to be searched ="); String look = br.readLine();

int flag = -1; while(st.hasMoreElements()) { if(look.equals(st.nextElement())) flag =1; } if(flag ==-1) {

System.out.println("the word not found"); }

else {

System.out.println("the word found"); }

} }

Output:

enter the string=

there are 7 days in a week

enter the word to be searched = are

the word found

Algorithm :

STEP 1 - START

STEP 2 - INPUT name, n STEP 3 - l=name.length()

STEP 4 - PRINT original string is "+name STEP 5 - IF i=0 THEN GOTO STEP 6 STEP 6 - char c1=name.charAt(i) STEP 7 - c=(int)c1

STEP 8 - IF n>0 THEN GOTO STEP 9 THERWISE GOTO STEP 12

STEP 9 - IF (c+n)<=90 THEN GOTO STEP 10 OTHERWISE GOTO STEP 11 STEP 10 - PRINT (char)(c+n)

STEP 11 - c=c+n;c=c%10,c=65+(c-1) & PRINT (char)(c)

STEP 12 - ELSE IF n<0 THEN GOTO STEP 13 OTHERWISE GOTO STEP 19 STEP 13 - n1=Math.abs(n)

STEP 14 - IF (c-n1) >=65 THEN GOTO STEP 15 OTHERWISE GOTO STEP 16 STEP 15 - DISPLAY (char) (c-n1)

STEP 16 - IF c>65 THEN GOTO STEP 17 OTHERWISE GOTO STEP 18 STEP 17 - c=c-65,

STEP 18 - c=n1 & PRINT (char)(90-(c-1)) STEP 19 - ELSE IF n==0

STEP 20 - DISPLAY "no change "+name STEP 21 - END

Solution :

class decode {

public void compute(String name,int n) {

int j,i,l,c=0,y,n1; l=name.length();

System.out.println("original string is "+name); for(i=0;i<l;i++) { char c1=name.charAt(i); try { c=(int)c1 ; }

catch(NumberFormatException e) {} if(n>0) { if((c+n)<=90) System.out.print((char)(c+n)); else { c=c+n;c=c%10; c=65+(c-1); System.out.print((char)(c)); } } else if(n<0) { n1=Math.abs(n); if((c-n1) >=65) System.out.print((char) (c-n1)); else { if(c>65) c=c-65; else c=n1; System.out.print((char)(90-(c-1))); } } else if (n==0) {

System.out.println("no change "+name); break; } } } }

Output:

original string is ABCDE n = 4

PROGRAM

21

-

To display the entered string in

alphabetical order

.

Algorithm :

STEP 1 - START

STEP 2 - str = "" , l = 0 STEP 3 - INPUT string str STEP 4 - l =str.length()

STEP 5 - FROM i=0 to i<l REPEAT STEP 6 STEP 6 - c[i] = str.charAt(i)

STEP 7 - FROM i=0 to i<l-1 REPEAT STEP 8 STEP 8 - FROM j=0 to i<l-1 REPEAT STEP 9

STEP 9 - temp =c[j], c[j] = c[j+1] , c[j+1] = temp STEP 10 - FROM i=0 to i<l REPEAT STEP 11

STEP 11 - PRINT c[i] STEP 12 - END

Solution:

char c[] = new char[100]; public Alpha() { str = ""; l =0; }

public void readword() throws IOException {

System.out.println("enter word - ");

BufferedReader br =new BufferedReader(new InputStreamReader(System.in));

l = str.length(); }

public void arrange() { int i,j; char temp; for(i=0;i<l;i++) { c[i]= str.charAt(i); } for(i=0;i<l-1;i++) { for(j=0;j<l-1-i;j++) { if(c[j] > c[j+1]) { temp = c[j]; c[j] = c[j+1]; c[j+1] = temp; } } } }

public void display() { System.out.println(); for(int i=0;i<l;i++) { System.out.print(c[i]); } }

public static void main(String args[]) throws IOException {

Alpha obj = new Alpha(); obj.readword(); obj.arrange(); obj.display(); } }

Output:

enter word - window dinoww

PROGRAM 2

2

-

To

create a string and count

number of vowels and consonants

.

Algorithm :

STEP 1 - START

STEP 2 - a = "Computer Applications" STEP 3 - z = a.length()

STEP 4 - x= 0 , b= 0

STEP 5 - FROM y =0 to y<z REPEAT STEP 6

STEP 6 - IF (a.charAt(y)=='a'||a.charAt(y)=='e'||a.charAt(y)=='i'|| a.charAt(y)=='o'||

a.charAt(y)=='u') THEN x =x +1 OTHERWISE b = b+1 STEP 7 - PRINT x STEP 8 - PRINT b STEP 9 - END

Solution:

public static void main(String args[]) {

String a="Computer Applications";//initialising string int z=a.length(),y,x=0,b=0;

for(y=0;y<z;y++)//loop for counting number of vowels { if(a.charAt(y)=='a'||a.charAt(y)=='e'||a.charAt(y)=='i'||a.charAt(y)=='o'|| a.charAt(y)=='u') x++; else b++; }

System.out.println("Number of vowels in string ="+x); System.out.println("Number of consonants in string ="+b); }

}

Output:

Number of vowels in string =7

Number of consonants in string =12

PROGRAM

23

-

To

create a string and count

number of

words

.

Algorithm :

STEP 1 - START

STEP 2 - a = "Computer Applications" STEP 3 - z = a.length()

STEP 4 - x= 0

STEP 5 - FROM y =0 to y<z REPEAT STEP 6 STEP 6 - IF (a.charAt(y)==' ' ) then x =x+1

STEP 7 - PRINT "Number of words in string ="+(x+1) STEP 8 - END

Solution:

class p45 {

public static void main(String args[]) {

String a="Computer Applications"; //initialising string System.out.println("The string is -"+a);

int z=a.length(),y,x=0;

for(y=0;y<z;y++)//loop for counting number of spaces {

if(a.charAt(y)==' ') x=x+1;

}

System.out.println("Number of words in string ="+(x+1)); }

Output:

The string is -Computer Applications Number of words in string =2

PROGRAM

24

-

To

create a string and

replace all

vowels with *

.

Algorithm :

STEP 1 - START

STEP 2 - a = "Computer Applications" STEP 3 - x= 0

STEP 4 - FROM z =0 to z<a.length() REPEAT STEP 5

STEP 5 - if(a.charAt(z)=='a'||a.charAt(z)=='e'||a.charAt(z)=='i'|| a.charAt(z)=='o'||

a.charAt(z)=='u') THEN a.setCharAt(z,'*') STEP 6 - PRINT "New String -"+a

STEP 7 - END

Solution:

import java.io.*; class p48

{

public static void main(String args[]) {

StringBuffer a=new StringBuffer("Computer Applications"); System.out.println("Original String -"+a);

int z=0;

for(z=0;z<a.length();z++)//loop for replacing vowels with "*" {

if(a.charAt(z)=='a'||a.charAt(z)=='e'||a.charAt(z)=='i'||a.charAt(z)=='o'|| a.charAt(z)=='u')

a.setCharAt(z,'*'); }

System.out.println("New String -"+a); }

}

Output:

New String -C*mp*t*r Appl*c*t**ns

PROGRAM 2

5

-

To

create a double-dimensional

array of 4*4 subscripts.

Algorithm :

STEP 1- START STEP 2- INPUT a[]

STEP 3- FROM x =0 to x<3 REPEAT STEP 4 STEP 4- FROM y =0 to y<3 REPEAT STEP 5 STEP 5- PRINT (a[x][y]+" "

STEP 6- END

Solution:

class p31 {

public static void main(String args[])throws IOException { int a[][]=new int[3][3], x,y,z;

BufferedReader aa=new BufferedReader(new InputStreamReader(System.in));

System.out.println("Enter the array");

for(x=0;x<3;x++)//loop for reading the array { for(y=0;y<3;y++) { z=Integer.parseInt(aa.readLine()); a[x][y]=z; } } System.out.println("Array -");

for(x=0;x<3;x++)//loop for printing the array { for(y=0;y<3;y++) { System.out.print(a[x][y]+" "); } System.out.print("\n"); } } }

Output:

Enter the array 1 2 3 4 5 6 7 8 9 Array -1 2 3 4 5 6 7 8 9

PROGRAM 2

6

-

To

generate sum of all elements of

a double dimensional array of 5*5 subscripts

Algorithm :

STEP 1 - START STEP 2 - INPUT a[]

STEP 3 - FROM x =0 to x<5 REPEAT STEP 4 STEP 4 - FROM y =0 to y<5 REPEAT STEP 5 STEP 5 - PRINT (a[x][y]+" "

STEP 6 - FROM x =0 to x<5 REPEAT STEP 7 STEP 7 - FROM y =0 to y<5 REPEAT STEP 8 STEP 8 - Sum=Sum+a[x][y]

STEP 9 - PRINT Sum STEP10 - END

Solution:

class p33 {

public static void main(String args[])throws IOException { int a[][]=new int[5][5];

BufferedReader aa=new BufferedReader(new InputStreamReader(System.in));

int x,y,z,Sum=0;

for(x=0;x<5;x++)//loop for reading array { for(y=0;y<5;y++) { z=Integer.parseInt(aa.readLine()); a[x][y]=z; } } System.out.println("Array -");

for(x=0;x<5;x++)//loop for printing array { for(y=0;y<5;y++) { System.out.print(a[x][y]+" "); } System.out.print("\n"); }

for(x=0;x<5;x++)//loop for printing sum of array elements { for(y=0;y<5;y++) { Sum=Sum+a[x][y]; } }

System.out.println("Sum of Array elements="+Sum); }

}

Output:

Enter the array 1 2 3 4 5 6 7 8 8 9 0 9 8 7 6 5

4 3 2 1 2 3 4 5 6 Array -1 2 3 4 5 6 7 8 8 9 0 9 8 7 6 5 4 3 2 1 2 3 4 5 6

Sum of Array elements=118

PROGRAM 2

7

-

To

generate product of two arrays

of 5 subscripts as a third array.

Algorithm :

STEP 1 - START

STEP 2 - INPUT a[] , b[]

STEP 3 - FROM i =0 to i<5 REPEAT STEP 4 and STEP 5 STEP 4 - c[i] = a[i] * b[i]

STEP 5 - PRINT c[i] STEP 6 - END

Solution:

class p35 {

public static void y(int a[],int b[]) { int c[]=new int[5];

int i;

System.out.println("Product of two arrays is-");

for(i=0;i<5;i++)//loop for finding product of the two arrays { c[i]=a[i]*b[i];

System.out.print(+c[i]+" "); }

}

Output:

Product of two arrays is-4 6 21 32 25

PROGRAM 2

8

-

To

find sum of each column of a

double dimensional array.

Algorithm :

STEP 1 - START STEP 2 - INPUT a[]

STEP 3 - FROM x =0 to x<4 REPEAT STEP 4 STEP 4 - FROM y =0 to y<4 REPEAT STEP 5 STEP 5 - PRINT (a[x][y]+" "

STEP 6 - FROM x =0 to x<4 REPEAT STEP 7 , STEP 9 and STEP 10 STEP 7 - FROM y =0 to y<4 REPEAT STEP 8

STEP 8 - Sum=Sum+a[x][y] , STEP 9 - PRINT Sum

STEP 10 - Sum = 0 STEP11 - END

Solution:

class p39 {

public static void main(String args[])throws IOException {

int a[][]=new int[4][4];

BufferedReader aa=new BufferedReader(new InputStreamReader(System.in));

int x,y,z,Sum=0;

System.out.println("Enter the array");//reading array for(x=0;x<4;x++) { for(y=0;y<4;y++) { z=Integer.parseInt(aa.readLine()); a[x][y]=z; } }

for(x=0;x<4;x++) { for(y=0;y<4;y++) { System.out.print(a[x][y]+" "); } System.out.print("\n"); } for(y=0;y<4;y++) { for(x=0;x<4;x++) { Sum=Sum+a[x][y]; }

System.out.println("Sum of column "+(y+1)+" is "+Sum);//printing sum of Sum=0; //column

} } }

Output:

Enter the array 1 2 3 4 5 6 7 8 9 0 9 8 7 6 5 4 Array -1 2 3 4 5 6 7 8 9 0 9 8 7 6 5 4 Sum of column 1 is 22 Sum of column 2 is 14

Sum of column 3 is 24 Sum of column 4 is 24

PROGRAM 2

9

-

To

find sum of diagonal of a double

dimensional array of 4*4 subscripts.

Algorithm :

STEP 1- START STEP 2- INPUT a[]

STEP 3- FROM x =0 to x<4 REPEAT STEP 4 STEP 4- FROM y =0 to y<4 REPEAT STEP 5 STEP 5- PRINT (a[x][y]+" "

STEP 6- FROM x =0 to x<4 REPEAT STEP 7 STEP 7 - Sum=Sum+a[x][y] , y=y+1

STEP 9- PRINT Sum STEP 10 - Sum = 0 STEP11- END

Solution:

class p40 {

public static void main(String args[])throws IOException {

int a[][]=new int[4][4];

BufferedReader aa=new BufferedReader(new InputStreamReader(System.in));

int x,y,z,Sum=0;

System.out.println("Enter the array"); for(x=0;x<4;x++) { for(y=0;y<4;y++) { z=Integer.parseInt(aa.readLine()); a[x][y]=z; } } System.out.println("Array -"); for(x=0;x<4;x++) { for(y=0;y<4;y++)

{ System.out.print(a[x][y]+" "); } System.out.print("\n"); } y=0;

for(x=0;x<4;x++)//loop for finding sum of diagonal {

Sum=Sum+a[x][y]; y=y+1;

}

System.out.println("Sum of diagonal is " +Sum); Sum=0;

} }

Output:

Enter the array 1 2 3 4 5 6 7 8 9 0 9 8 7 6 5 4 Array -1 2 3 4 5 6 7 8 9 0 9 8 7 6 5 4 Sum of diagonal is 20

PROGRAM

30

-

To

calculate the commission of a

salesman as per the following data:

Sales Commission >=100000 25% of sales 80000-99999 22.5% of sales 60000-79999 20% of sales 40000-59999 15% of sales <40000 12.5% of sales

Algorithm :

STEP 3 - IF (sales>=100000) THEN comm=0.25 *sales OTHERWISE GOTO

STEP 4

STEP 4 - IF (sales>=80000) THEN comm=0.225*sales OTHERWISE GOTO

STEP 5

STEP 5 - IF (sales>=60000) THEN comm=0.2 *sales OTHERWISE GOTO

STEP 6

STEP 6 - IF (sales>=40000) THEN comm=0.15 *sales OTHERWISE GOTO

STEP 7

STEP 7 - comm=0.125*sales

STEP 8 - PRINT "Commission of the employee="+comm STEP 9 - END

Solution:

class p14 {

public static void main(String args[])throws IOException {

double sales,comm;

BufferedReader aa=new BufferedReader(new InputStreamReader(System.in));

System.out.println(“Enter sales”);

sales=Double.parseDouble(aa.readLine());//reading sales from the keyboard if(sales>=100000)

comm=0.25*sales; else

if(sales>=80000) comm=0.225*sales;

else if(sales>=60000) comm=0.2*sales; else if(sales>=40000) comm=0.15*sales; else comm=0.125*sales;

System.out.println("Commission of the employee="+comm); } }

Output: