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
PROGRAM 20 - To decode the entered string
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
decoded string is EFGHI
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:
import java.io.*;
class Alpha {
String str;
int l;
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));
str = br.readLine();
l = str.length();
public static void main(String args[]) throws IOException {
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:
class p42 {
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:
Original String -Computer Applications
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;
System.out.println("Enter the array");
for(x=0;x<5;x++)//loop for reading array
for(x=0;x<5;x++)//loop for printing array {
for(x=0;x<5;x++)//loop for printing sum of array elements {
System.out.println("Sum of Array elements="+Sum);
}
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;
} }
System.out.println("Array -");//printing the array in matrix form
for(x=0;x<4;x++)
System.out.println("Sum of column "+(y+1)+" is "+Sum);//printing sum of Sum=0; //column
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 1 - START STEP 2 - INPUT sales
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:
Enter sales 60000
Commission of employee=12000