Question

cis 112 (java reqire to use JGRASP)

Searching and Sorting Assignment This is a non-object oriented assignment. This assignment is due at the start of class on July 16, Create a program that populates and array with 100 random integers between 0 and 99. Then print out the array. Then prompt the user for a number, and do a sequential search on the unsorted array and return whether or not the number was in the array. Then sort the array using the sort method of your choosing. Make sure you document your method: Print out sufficient local variables for each iteration of the sorting algorithm so that you can see how it is working Print out the contents of the array to show the sort worked. Then prompt the user for another number. The do a binary search on the sorted array to return whether or not the number is in the array Remember to adequately document your program. You can copy code from the Geeks website, but you must add the additional things mentioned above, plus comments!!

Answer 1

PROGRAM CODE

import java.util.Random;
import java.util.Scanner;

/**
* The following program has an array of integers filled with random numbers between 0 and 99
* It will first ask user for a number to search in array and perform sequential search
* Next, it will sort the array using Bubble Sort
* Next, it will ask for another number and perform binary search on that sorted array to find number
*
* @author Anonymous
* @version 1.0
* @since 2020-07-28
*/

public class SearchArray {

public static void main(String[] args) {

// Declaring an integer array of 100 elements

int[] array = new int[100];

// Populating array with random elements between 0 and 99

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

array[i] = new Random().nextInt(99);
}

// Printing out the numbers in array using print method

print(array);

// Asking user for a number to search in array sequentially

Scanner input = new Scanner(System.in); // object of Scanner class to take input

System.out.print("\nEnter a number to search for in array: ");
int number = input.nextInt();

// Searching Sequentially for that number in array and if found, show it is found

boolean isFound = false; // Variable to determine whether value has found or not

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

if(number == array[i]){

System.out.println("The number is in the array.");
isFound = true;
break;
}
}

if(!isFound)
System.out.println("The number is not in the array.");

// Calling sort function and then printing array to see it is sorted or not

sort(array, 100);
System.out.println("\nThe sorted array is");
print(array);

// Performing Binary search by calling method of binarySearch

System.out.print("\nEnter a number to search for in array: ");
number = input.nextInt();

binarySearch(array, 100, number);
}

/**
* This method will print the elements of the array
* @param array The array of integers whose elements are to be printed
*/

public static void print(int[] array) {

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

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

// if(i%10 == 0 && i>0) // After every 10 elements print new line
// System.out.println();
}
System.out.println();
}

/**
* This method will sort the values of the passed array in ascending order using Bubble Sort
* @param array array of integers
* @param size size of array
*/

public static void sort(int[] array, int size) {

for (int i=0; i<size; i++) {

for (int j=0; j<size-1; j++) {

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

int temp = array[j];
array[j] = array[j+1];
array[j+1] = temp;
}

// Showing local variables with each iteration of algorithm, uncomment below to see iteration
//
// System.out.println("Iteration #" + j+1 + " of Pass#" + i+1);
// print(array);
}
}
}

/**
* This method will perform binary search on array
* @param array The array of integers
* @param size The size of array
* @param number The number to be find in array
*/

public static void binarySearch(int[] array,int size, int number) {

int first = array[0];
int end = array[size - 1];
int middle = (first + end) / 2;

boolean isFound = false;

while (first <= end) {

if(array[middle] == number) {
System.out.println("The number is in the array.");
isFound = true;
break;
}
else if(array[middle] < number) {
first = middle + 1; // proceed to next number
}
else{
end = middle - 1;
}

middle = (first + end) / 2;
}

if(!isFound) {

System.out.println("The number is not in the array.");
}
}
}

SAMPLE OUTPUT

SearchArray "C:\Program Files\Java\jdk1.8.0_111\bin\java.exe" ... 78 79 2 80 11 2 13 22 74 0 24 32 47 54 7 13 37 67 41 42 8 36 94 95 85 94 23 34 0 12 53 5 70 67 39 68 68 65 71 76 4 68 23 23 75 32 92 44 79 65 11 88 51 14 59 Enter a number to search for in array: 80 The number is in the array. The sorted array is 0 0 2 2 3 4 4 5 6 7 7 8 11 11 12 12 12 13 13 14 14 15 19 19 20 22 22 23 23 23 23 24 25 26 27 29 30 31 31 32 32 32 33 34 36 36 37 39 39 40 40 41 41 41 41 42 Enter a number to search for in array: 100 The number is not in the array. Process finished with exit code 0

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