What could be one possible reason where the recipient is not guaranteed that the data being streamed will not get interrupted?
1 answer:
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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
