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Elena-2011 [213]
2 years ago
11

Evaluating sorts given the following array: 41, 32, 5, 8, 7, 50, 11 show what the array looks like after the first swap of a bub

ble sort in a scending order.
Computers and Technology
1 answer:
irga5000 [103]2 years ago
8 0

The way that  the array will looks like after the first swap of a bubble sort in ascending order is   {5,32,41, 8,7,50,11}.

<h3>Who do you Write sort code in JAVA?</h3>

It will be:

class SSort

{

  void sort(int array[])

  {

      int n = array.length;

      for (int i = 0; i < 2; i++)

      {

          int minimum_index = i;

          for (int j = i+1; j < n; j++)

              if (array[j] < array[minimum_index])

                  minimum_index = j;

           int temp = array[minimum_index];

          array[minimum_index] = array[i];

          array[i] = temp;

      }

  }

   void printArray(int array[])

  {

      int n = array.length;

      for (int i=0; i<n; ++i)

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

      System.out.println();

  }

  public static void main(String args[])

  {

      SSort s = new SSort();

      int array[] = {41, 32, 5, 8, 7, 50, 11};

      s.sort(array);

      System.out.println("After first two iteration of Selection sort:");

      s.printArray(array);

  }

}

Therefore, The way that  the array will looks like after the first swap of a bubble sort in ascending order is   {5,32,41, 8,7,50,11}.

Learn more about array  from

brainly.com/question/26104158

#SPJ1

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Answer:

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Answer:

  1. import java.util.Scanner;
  2. public class Main {
  3.    public static void main(String[] args) {
  4.        int num[] = new int[51];
  5.        Scanner input = new Scanner(System.in);
  6.        System.out.print("Number of input: ");
  7.        int limit = input.nextInt();
  8.        for(int i=0; i < limit; i++){
  9.            System.out.print("Input a number (1-50): ");
  10.            int k = input.nextInt();
  11.            num[k]++;
  12.        }
  13.        for(int j=1; j < 51; j++){
  14.            if(num[j] > 0){
  15.                System.out.println("Number of occurrence of " + j + ": " + num[j]);
  16.            }
  17.        }
  18.    }
  19. }

Explanation:

The solution is written in Java.

Firstly, create an integer array with size 51. The array will have 51 items with initial value 0 each (Line 5).

Create a Scanner object and get user entry the number of input (Line 6-7).

Use the input number as the limit to control the number of the for loop iteration to repeatedly get integer input from user (Line 9-13). Whenever user input an integer, use that integer, k, as the index to address the corresponding items in the array and increment it by one (LINE 11-12).

At last, create another for loop to iterate through each item in the array and check if there is any item with value above zero (this means with occurrence at least one). If so, print the item value as number of occurrence (Line 14-17).

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