Case-Based Critical Thinking QuestionsCase 9-1Melissa, a computer science engineering student, is learning the basics of progr
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
Hi, you haven't provided the programing language in which you need the code, I'll explain how to do it using Python, and you can follow the same logic to make a program in the programing language that you need.
Answer:
import math
def rectangle(perimeter, area):
l1_1 = (perimeter+math.sqrt((perimeter**2)-(16*area)))/4
l1_2 = (perimeter-math.sqrt((perimeter**2)-(16*area)))/4
l2_1 = area/l1_1
l2_2 = area/l1_2
print(l1_1,l2_1)
print(l1_2,l2_2)
if l1_1.is_integer() and l2_1.is_integer() and l1_1>0 and l2_1>0:
return(int(max(l1_1,l2_1)))
elif l1_2.is_integer() and l2_2.is_integer() and l1_2>0 and l2_2>0:
return(int(max(l1_2,l2_2)))
else:
return(None)
Explanation:
- We import math to make basic operations
- We define the rectangle function that receives perimeter and area
- We calculate one of the sides (l1_1) of the rectangle using the quadratic equation to solve 2h^2 - ph + 2a = 0
- We calculate the second root of the quadratic equation for the same side (l1_2)
- We calculate the second side of the rectangle using the first root on w = a/h
- We calculate the second side of the rectangle using the second root on w= a/h
- We verify that each component of the first result (l1_1, l2_1) is an integer (using the build-in method .is_integer) and greater than 0, if True we return the maximum value between them (using the max function) as w
- If the first pair of sides evaluate to False we check the second root of the equation and if they meet the specification we return the max value
- if all the if statements evaluate to false we return None to indicate that not positive or integer sides were found
