Use C++ and solve the problem without a sorting algorithm.here is the code I have:#include <iostream>#include <bits/stdc++.h>using namespace std;struct Node {int data;Node* next;};void insert(Node** root, int item){Node* temp = new Node;Node* ptr;temp->data = item;temp->next = NULL;if (*root == NULL)*root = temp;else{ptr = *root;while (ptr->next != NULL)ptr = ptr->next;ptr->next = temp;}}void display(Node* root){while (root != NULL){cout << root->data << " ";root = root->next;}cout<<endl;}Node *arrayToList(int arr[], int n){Node *root = NULL;for (int i = 0; i < n; i++)insert(&root, arr[i]);return root;}int main(){int arr[]={4,3,6,1,10,9,5,8,7,2};int n = sizeof(arr) / sizeof(arr[0]);for(int i = 1; i <= n; i++){sort(arr, arr+i);Node* root = arrayToList(arr, i);display(root);}return 0; The array contains all 10 different integers. Store them in a linked list in order and print them out.**Block of Code:**```cint main(){ int A[10] = { 4, 3, 6, 1, 10, 9, 5, 8, 7, 2 }; return 0;}```**Diagram Explanation:**The diagram on the left is a visual representation showing different stages of a linked list creation and ordering process. It appears to depict an array of integers with lines representing steps in sorting or linking. Each line presents integers, successively grouped to show what might be sub-steps or intermediate linked list states leading to the final sequence: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. ### Code Example: Working with Rational Numbers```cppint main(){ Rational r1, r2; char answer = 'y'; while (answer == 'y') { cout << "Enter the first fraction (e.g. 3/4) : "; cin >> r1; cout << "Enter the second fraction (e.g. 3/4) : "; cin >> r2; cout << "r1 = " << r1 << endl; cout << "r2 = " << r2 << endl; cout << "r1 + r2 = " << r1 + r2 << endl; cout << "r1 - r2 = " << r1 - r2 << endl; cout << "r1 * r2 = " << r1 * r2 << endl; cout << "r1 / r2 = " << r1 / r2 << endl; cout << "r1 == r2 = " << (r1 == r2) << endl; cout << "r1 != r2 = " << (r1 != r2) << endl; cout << "r1 < r2 = " << (r1 < r2) << endl; cout << "r1 <= r2 = " << (r1 <= r2) << endl; cout << "r1 > r2 = " << (r1 > r2) << endl; cout << "r1 >= r2 = " << (r1 >= r2) << endl; cout << "Again (y/n)? "; cin >> answer; } return 0;}```### Output Explanation1. **User Input and Calculations:** - Users are prompted to input two fractions in the form `a/b`. - The program performs various arithmetic and comparative operations on these fractions: - Addition (`r1 + r2`) - Subtraction (`r1 - r2`) - Multiplication (`r1 * r2`) - Division (`r1 / r2`) - Checking equality (`r1 == r2`) - Checking inequality (`r1 != r2`) - Comparing sizes (`r1 < r2`, `r1 <= r2`, `r1 > r2`, `r

We need to use C++ and solve the given problem without a sorting algorithm.

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Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

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Use C++ and solve the problem without a sorting algorithm.

here is the code I have:

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
Node* next;
};
void insert(Node** root, int item)
{
Node* temp = new Node;
Node* ptr;
temp->data = item;
temp->next = NULL;
if (*root == NULL)
*root = temp;
else
{
ptr = *root;
while (ptr->next != NULL)
ptr = ptr->next;
ptr->next = temp;
}
}
void display(Node* root)
{
while (root != NULL)
{
cout << root->data << " ";
root = root->next;
}
cout<<endl;
}
Node *arrayToList(int arr[], int n)
{
Node *root = NULL;
for (int i = 0; i < n; i++)
insert(&root, arr[i]);
return root;
}
int main()
{
int arr[]={4,3,6,1,10,9,5,8,7,2};
int n = sizeof(arr) / sizeof(arr[0]);

for(int i = 1; i <= n; i++)
{
sort(arr, arr+i);
Node* root = arrayToList(arr, i);
display(root);}
return 0;

The array contains all 10 different integers. Store them in a linked list in order and print them out.

**Block of Code:**

```c
int main()
{
int A[10] = { 4, 3, 6, 1, 10, 9, 5, 8, 7, 2 };

return 0;
}
```

**Diagram Explanation:**

The diagram on the left is a visual representation showing different stages of a linked list creation and ordering process. It appears to depict an array of integers with lines representing steps in sorting or linking. Each line presents integers, successively grouped to show what might be sub-steps or intermediate linked list states leading to the final sequence: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

### Code Example: Working with Rational Numbers

```cpp
int main()
{
Rational r1, r2;
char answer = 'y';

while (answer == 'y') {
cout << "Enter the first fraction (e.g. 3/4) : "; cin >> r1;
cout << "Enter the second fraction (e.g. 3/4) : "; cin >> r2;

cout << "r1 = " << r1 << endl;
cout << "r2 = " << r2 << endl;
cout << "r1 + r2 = " << r1 + r2 << endl;
cout << "r1 - r2 = " << r1 - r2 << endl;
cout << "r1 * r2 = " << r1 * r2 << endl;
cout << "r1 / r2 = " << r1 / r2 << endl;
cout << "r1 == r2 = " << (r1 == r2) << endl;
cout << "r1 != r2 = " << (r1 != r2) << endl;
cout << "r1 < r2 = " << (r1 < r2) << endl;
cout << "r1 <= r2 = " << (r1 <= r2) << endl;
cout << "r1 > r2 = " << (r1 > r2) << endl;
cout << "r1 >= r2 = " << (r1 >= r2) << endl;
cout << "Again (y/n)? ";
cin >> answer;
}

return 0;
}
```

### Output Explanation

1. **User Input and Calculations:**
- Users are prompted to input two fractions in the form `a/b`.
- The program performs various arithmetic and comparative operations on these fractions:
- Addition (`r1 + r2`)
- Subtraction (`r1 - r2`)
- Multiplication (`r1 * r2`)
- Division (`r1 / r2`)
- Checking equality (`r1 == r2`)
- Checking inequality (`r1 != r2`)
- Comparing sizes (`r1 < r2`, `r1 <= r2`, `r1 > r2`, `r

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