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Degger [83]
3 years ago
14

You will be given a string, containing both uppercase and lowercase alphabets(numbers are not allowed).

Computers and Technology
1 answer:
murzikaleks [220]3 years ago
6 0

Answer:

The Java code is given below with appropriate comments

Explanation:

import java.util.*;

class Main {

   static Set<String> set = new HashSet();

   static void printPerms(char ch[], int ind){

       //If end of string is reached, add it to set

       if(ind==ch.length){

           set.add(new String(ch));

           return;

       }

       for(int i=ind;i<ch.length;i++){

           //Only swap if lower case

           if((ch[i]>='a' && ch[i]<='z')&&((ch[ind]>='a' && ch[ind]<='z'))){

               char t  = ch[i];

               ch[i] = ch[ind];

               ch[ind] = t;

           }

           printPerms(ch,ind+1);

           if((ch[i]>='a' && ch[i]<='z')&&((ch[ind]>='a' && ch[ind]<='z'))){

               char t  = ch[i];

               ch[i] = ch[ind];

               ch[ind] = t;

           }

       }

   }

   public static void main(String[] args) {

       printPerms("aBbCc".toCharArray(),0);

       System.out.println(set);

   }

}

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Answer:

Option B is the correct answer for the following question.

Explanation:

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g Write a C program that prompts the user for a string of, up to 50, characters and numbers. It can accept white spaces and tabs
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Answer:

Following are the code to this question:

#include <stdio.h> //defining header file

#include <string.h>//defining header file

int main()//defining main method

{

char name[50]; //defining character variable

int A=0,E=0,I=0,O=0,U=0,value=0,i,l; //defining integer variables

printf("Enter string value: ");//print message

gets(name);//input value by gets method

l=strlen(name);//defining l variable to store string length

for(i=0; i<50; i++) //defining loop to count vowels value

{

if(name[i] == 'A' || name[i] == 'a') //defining if condition count A value A++;//increment value by 1

else if(name[i] == 'E' || name[i] == 'e')//defining else if condition count E value

E++;//increment value by 1

else if(name[i] == 'I' || name[i] == 'i') //defining else if condition count I value

I++;//increment value by 1

else if(name[i] == 'O' || name[i]== 'o') //defining else if condition count O value

O++;//increment value by 1

else if(name[i] == 'U' || name[i] == 'u')//defining else if condition count U value

U++;//increment value by 1

else if(name[i]>= '0' && name[i] <= '9')//defining else if condition count digits value

value++;//increment value by 1 }

printf("A: %d\n",A);//print A value

printf("E: %d\n",E); //print E value

printf("I: %d\n",I);//print I value

printf("O: %d\n",O);//print O value

printf("U: %d\n",U);//print U value

printf("value: %d",value);//print digits value

return 0;

}

Output:

please find the attachment.

Explanation:

  • In the above code, a char array "name" and integer variables "A, E, I, O, U, i, value and l" declared, in which name array store 50 character value, which uses the gets function to take string value.after input value strlen method is used to store the length of the input string in l variable.
  • In the next step, for loop is declared that uses multiple conditional statements to counts vowels and the values of the digits,if it is more then one it increments its value and at the last print method is used to print its count's value.

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