Question

JAVA SHOW YOUR OUTPUT

package exampletodo;

import java.util.Arrays;

import stdlib.*;

/**

* Edit the sections marked TODO

*

* Unless specified otherwise, you must not change the declaration of any

* method.

*/

public class example {

       /**

       * valRange returns the difference between the maximum and minimum values in the

       * array; Max-Min. Precondition: the array is nonempty. Your solution must go

       * through the array at most once.

       *

       * Here are some examples (using "==" informally):

       *

       * <pre>

       *    0 == valRange (new double[] { -7 })

       *    10 == valRange (new double[] { 1, 7, 8, 11 })

       *    10 == valRange (new double[] { 11, 7, 8, 1 })

       *    18 == valRange (new double[] { 1, -4, -7, 7, 8, 11 })

       *    24 == valRange (new double[] { -13, -4, -7, 7, 8, 11 })

       *

       * The code below is a stub version, you should replace the line of code

       * labeled TODO with code that achieves the above specification

       * </pre>

       */

       public static double valRange(double[] list) {

             return -1; // TODO 1: fix this code

       }

       /**

       * posOfLargestElementLtOeT returns the position of the largest element in the

       * array that is less than or equal to the limit parameter if all values are

       * greater than limit, return -1;

       *

       * Precondition: the array is nonempty and all elements are unique. Your

       * solution must go through the array exactly once.

       *

       * <pre>

       *   0 == posOfLargestElementLtOeT(3, new double[] { -7 })                      // value:-7 is in pos 0

       *   5 == posOfLargestElementLtOeT(3, new double[] { 11, -4, -7, 7, 8, 1 }),    // value:1 is in pos 5

       * -1 == posOfLargestElementLtOeT(-7, new double[] { 1, -4, -5, 7, 8, 11 }),   // all elements are > -7

       *

       * The code below is a stub version, you should replace the line of code

       * labeled TODO with code that achieves the above specification

       * </pre>

       */

       public static int posOfLargestElementLtOeT(double limit, double[] list) {

             return -2; // TODO 2: fix this code

       }

       /**

       * isPerfectNumber determines (true or false) if a given number is a 'Perfect

       * Number'

       *

       * A perfect number is one that is equal to the sum of its proper divisors.

       * Example 1: 6; the proper divisors are 1, 2, 3 ; 1+ 2 + 3 is 6, so 6 IS a

       * perfect number Example 2: 15; the proper divisors are 1, 3, 5 ; 1 + 3 + 5 is

       * 9, so 15 IS NOT a perfect number Example 3: 28; the proper divisors are 1, 2,

       * 4, 7, 14; 1 + 2 + 4 + 7 + 14 is 28, so 28 IS a perfect number

       *

       * Precondition: number is a positive integer

       *

       * The code below is a stub version, you should replace the line of code labeled

       * TODO with code that achieves the above specification

       *

       * Hint: find the sum of the proper divisors

       */

       public static boolean isPerfectNumber(int number) {

             return false; // TODO 3: fix this code

       }

Answer 1

**CODE START**

import java.util.Arrays;

public class example
{

   public static double valRange(double[] list) {
       double smallest = list[0];
     
       //Find smallest element by traversing through the list
       for(int i=0;i<list.length;i++){
          if(smallest > list[i])
              smallest = list[i];
       }
     
       double largest = list[0];
     
       //Find largest element by traversing through the list
       for(int i=0;i<list.length;i++){
          if(largest < list[i])
              largest = list[i];
       }
   
        //return difference of largest and smallest
       return (largest-smallest);
    }

    public static int posOfLargestElementLtOeT(double limit, double[] list) {
         
          //Let the index of largest element be "index"
          //initially no largest element so index = -1
        int index = -1;
      
        for(int i=0;i<list.length;i++){
            //If current element is less than limit, then proceed
           if(list[i] <= limit){
                //If there is no largest element till now, make this one largest
                //Otherwise compare it with current largest element
               if(index==-1){
                   index = i;
               }
               else if(list[index] < list[i]){
                   index = i;
               }
           }
        }
  
        return index;
    }

    public static boolean isPerfectNumber(int number) {
  
        int sum = 0;
      
        //Taking sum of all its divisors
        for(int i=1;i<number;i++){
           if(number%i==0)
               sum+=i;
        }
      
        //If sum==number return true, else return false
        if(sum==number)
           return true;
        return false;
    }  
     
     
   public static void main(String[] args) {
        System.out.println(valRange (new double[] { -7 }));
        System.out.println(valRange (new double[] { 1, 7, 8, 11 }));
        System.out.println(valRange (new double[] { 11, 7, 8, 1 }));
        System.out.println(valRange (new double[] { 1, -4, -7, 7, 8, 11 }));
        System.out.println(valRange (new double[] { -13, -4, -7, 7, 8, 11 }));
      
        System.out.println(posOfLargestElementLtOeT(3, new double[] { -7 }));
        System.out.println(posOfLargestElementLtOeT(3, new double[] { 11, -4, -7, 7, 8, 1 }));
        System.out.println(posOfLargestElementLtOeT(-7, new double[] { 1, -4, -5, 7, 8, 11 }));
      
        System.out.println(isPerfectNumber(6));
        System.out.println(isPerfectNumber(15));
        System.out.println(isPerfectNumber(28));
   }
}

**CODE END**

OUTPUT

**CODE SNIPPET END**