Question

please answer all qustions use this input(6587)

Problem Description: Consider afour digit number corresponding to number of letters in your full name. Input this number in your program to compute the below: a) Count the number of digits in the input number: Suppose if the input number is 3561, output should be 4. b) Compute the sum of digits in the input number: Suppose if the input number is 3561, output should be 3+5+6+1 = 15. C) Reverse the digits of the given number: Suppose if the input number is 3561, output should be 1653. d) Count the number of odd numbered digits in the given number. Suppose if the input number is 3561, output should be 3 since there are three odd numbered digits (3,5, and 1). Include the below in the report: 1. Algorithm for each task Time complexity of each task. 3. Tracing of each task by considering the number formed out of your name as the input. Java program along with screenshot of the solution to all four tasks [Number should correspond to your full name). 30

Answer 1

Here is the solution to your problem, if you are satisfied with the answer please give an upvote, else do ask your doubts in the comments.

Main.java B saved 1 import java.util.Scanner; 2 class Main // a) static int countdigit(int n) int i=1; while(n/10/0) n/=10; it=1; return i; // b) static int sumdigit(int n) int sum=0; while(n>0) sum+=n%10; n/=10; return sum; Il c) Main.java B saved 28 // c) static int reversedigit(int n) int rn=0; while(n>0) rn=rn*10+n%10; n/=10; return rn; // d) static int odddigit(int n) int cnt=0; while(n>0) if((n%10)%2==1) cnt==1; n/=10; return cnt; public static void main(String[] args), Scanner sc=new Scanner(System.in); saved Main.java B int cnt=0; while(n>0) if((n%10)%2==1) cnt==1; n/=10; return cnt; public static void main(String[] args) Scanner sc=new Scanner(System.in); System.out.print("Enter the number: "); int n=sc.nextInt(); System.out.print(n+" contains "+countdigit(n)+" digit.\n"); System.out.print(n+" of digits of "+sumdigit(n)+" is %d.\n"); System.out.print("Reverse of "+n+" is "+reversedigit(n)+".\n"); System.out.print("count of odd digits of "+n+" is "+odddigit(n)+", digit.\n"); 67 }

import java.util.Scanner; class Main { // a) static int countdigit(int n) { int i=1; while(n/10>0) { n/=10; i+=1; } return i; } // b) static int sumdigit(int n) { int sum=0; while(n>0) { sum+=n%10; n/=10; } return sum; } // c) static int reversedigit(int n) { int rn=0; while(n>0) { rn=rn*10+n%10; n/=10; } return rn; } // d) static int odddigit(int n) { int cnt=0; while(n>0) { if((n%10)%2==1) cnt+=1; n/=10; } return cnt; } public static void main(String[] args) { Scanner sc=new Scanner(System.in); System.out.print("Enter the number: "); int n=sc.nextInt(); System.out.print(n+" contains "+countdigit(n)+" digit.\n"); System.out.print(n+" of digits of "+sumdigit(n)+" is %d.\n"); System.out.print("Reverse of "+n+" is "+reversedigit(n)+".\n"); System.out.print("count of odd digits of "+n+" is "+odddigit(n)+" digit.\n"); } }

> javac -classpath .:/run_dir/junit-4.12.jar:target/dependency/* -d. Main.java > java -classpath .:/run dir/junit-4.12.jar:target/dependency/+ Main Enter the number: 6587 6587 contains 4 digit. 6587 of digits of 26 is fd. Reverse of 6587 is 7856. count of odd digits of 6587 is 2 digit.

1. Algorithms

Algorithm countdigit(n): 1. set i=1, (as the lowest amount of digits count of any number) 2. while n/10>0, repeat steps 3 and 4 3. n=n/10 4. i=i+1 5 return i Algorithm sumdigit(n): 1. set sum=0, (as the lowest sum of digits of any number) 2. while n>0, repeat steps 3 and 4 3. sum=sum+n%10 4. n=n/10 5 return sum Algorithm reversedigit(n): 1. set rn=0 2. while n>0, repeat steps 3 and 4 3. rn=rn*10+n%10 4. n=n/10 5 return rn Algorithm odddigit(n): 1. set cnt=0 2. while n>0, repeat steps 3 and 4 3. if (n%10)%2==1, then cnt=cnt+1 4. n=n/10 5 return cnt

The complexity of all function is O(logn).