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romanna [79]
2 years ago
8

Use the conversion table below to help you answer the question.

Computers and Technology
1 answer:
andrew-mc [135]2 years ago
7 0
Correct answer is 11
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Explanation:

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Consider the following algorithms. Each algorithm operates on a list containing n elements, where n is a very large integer.
DochEvi [55]

We have that the appropriate statement will be

  • An algorithm that accesses only the first 10 elements in the list, regardless of the size of the list. Which of the algorithms run in reasonable time

III only.

Option B

From the question we are told

Consider the following <u>algorithms</u>. Each <u>algorithm</u> operates on a list containing n <em>elements</em>, where n is a very large <u>integer</u>.

I. An algorithm that accesses each <u>element</u> in the list twice.

II. An <em>algorithm </em>that accesses each <u>element </u>in the list n times.

III. An <u>algorithm</u> that accesses only the first 10 elements in the list, regardless of the size of the list. Which of the <em>algorithms </em>run in <em>reasonable </em>time?

<h3>Algorithm  </h3>

Generally In order to get <em>admission </em>to every thing in the list twice, the algorithm has to traverse the listing twice,

which leads to 2*n entry to operations.

And if the every factor is accessed n times, the listing will be traversed n time, which leads to n^2 get right of entry to operations.

If n is a very giant <em>integer</em>, each 2*n and n^2 are plenty larger.

So, there will be <em>solely </em>ten entry to operations and this algorithm will have a sensible jogging time.

Therefore

An algorithm that accesses only the first 10 elements in the list, regardless of the size of the list. Which of the algorithms run in reasonable time

III only.

Option B

For more information on  algorithm  visit

brainly.com/question/950632

4 0
2 years ago
Write a Java class to perform the following: 1. Write a method to search the following array using a linear search, ( target ele
Alina [70]

Answer:

Check the explanation

Explanation:

Linear search in JAVA:-

import java.util.Scanner;

class linearsearch

{

  public static void main(String args[])

  {

     int count, number, item, arr[];

     

     Scanner console = new Scanner(System.in);

     System.out.println("Enter numbers:");

     number = console.nextInt();

   

     arr = new int[number];

     System.out.println("Enter " + number + " ");

     

     for (count = 0; count < number; count++)

       arr[count] = console.nextInt();

     System.out.println("Enter search value:");

     item = console.nextInt();

     for (count = 0; count < number; count++)

     {

        if (arr[count] == item)

        {

          System.out.println(item+" present at "+(count+1));

         

          break;

        }

     }

     if (count == number)

       System.out.println(item + " doesn't found in array.");

  }

}

Kindly check the first attached image below for the code output.

Selection Sort in JAVA:-

public class selectionsort {

   public static void selectionsort(int[] array){

       for (int i = 0; i < array.length - 1; i++)

       {

           int ind = i;

           for (int j = i + 1; j < array.length; j++){

               if (array[j] < array[ind]){

                   ind = j;

               }

           }

           int smaller_number = array[ind];  

           array[ind] = array[i];

           array[i] = smaller_number;

       }

   }

     

   public static void main(String a[]){

       int[] arr = {9,94,4,2,43,18,32,12};

       System.out.println("Before Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

       System.out.println();

         

       selectionsort(arr);

       

       System.out.println("After Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

   }

}  

Kindly check the second attached image below for the code output.

Bubble Sort in JAVA:-

public class bubblesort {

   static void bubblesort(int[] array) {

       int num = array.length;

       int temp = 0;

        for(int i=0; i < num; i++){

                for(int j=1; j < (num-i); j++){

                         if(array[j-1] > array[j]){

                           

                                temp = array[j-1];

                                array[j-1] = array[j];

                                array[j] = temp;

                        }

                         

                }

        }

   }

   public static void main(String[] args) {

               int arr1[] ={3333,60,25,32,55,620,85};

               

               System.out.println("Before Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

               System.out.println();

                 

               bubblesort(arr1);

               

               System.out.println("After Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

 

       }

}  

Kindly check the third attached image below for the code output.

Binary search in JAVA:-

public class binarysearch {

  public int binarySearch(int[] array, int x) {

     return binarySearch(array, x, 0, array.length - 1);

  }

  private int binarySearch(int[ ] arr, int x,

        int lw, int hg) {

     if (lw > hg) return -1;

     int middle = (lw + hg)/2;

     if (arr[middle] == x) return middle;

     else if (arr[middle] < x)

        return binarySearch(arr, x, middle+1, hg);

     else

        return binarySearch(arr, x, lw, middle-1);

  }

  public static void main(String[] args) {

     binarysearch obj = new binarysearch();

     int[] ar =

       { 22, 18,12,14,36,59,74,98,41,23,

        34,50,45,49,31,53,74,56,57,80,

        61,68,37,12,58,79,904,56,99};

     for (int i = 0; i < ar.length; i++)

        System.out.print(obj.binarySearch(ar,

           ar[i]) + " ");

     System.out.println();

     System.out.print(obj.binarySearch(ar,19) +" ");

     System.out.print(obj.binarySearch(ar,25)+" ");

     System.out.print(obj.binarySearch(ar,82)+" ");

     System.out.print(obj.binarySearch(ar,19)+" ");

     System.out.println();

  }

}

Kindly check the fourth attached image below for the code output

7 0
3 years ago
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