Question

Homework 2: Sorting In the second homework you will implement two sorting algorithms of your choice. One sorting algorithm has O(n) complexity while the other has O(n.logn) complexity. The input is student data stored in a file. Student data has the format: Name ID GPA The key for sorting of the O(n) algorithm is the ID while the key for sorting of the O(n.logn) is GPA. Your code should have all the necessary structures and functions to perform the required operations including:

Need some help on this, thanks. Must be done in C language.

Answer 1

solution:

//C program

#include<stdio.h>

struct student{

char name [20];

int ID;

double GPA;

};

void print Data(struct student st[] , int n) {

int i;

for( i=0;i<n;i++){
       printf("Student %d\n",i+1);
      
       printf("Name : %s\n",st[i].name);
       printf("ID : %d\n",st[i].ID);
       printf("GPA : %.2lf\n",st[i].GPA);
       printf("-------------------------------\n\n");
   }
}

void bubbleSort(struct student a[],int n){
   int i,j;
   for(i=n-1;i>=0;i--){
       for(j=0;j<i;j++){
           if(a[j].ID>a[j+1].ID){
               struct student temp=a[j];
               a[j]=a[j+1];
               a[j+1]=temp;
           }
       }
   }
}

void merge(struct student a[],struct student temp[],int l,int m,int r){
   int temp_pos=l,size=r-l+1,left_end=m-1;
   while(l<=left_end&&m<=r){
       if(a[l].GPA<a[m].GPA)temp[temp_pos++]=a[l++];
       else temp[temp_pos++]=a[m++];
   }
   while(l<=left_end)temp[temp_pos++]=a[l++];
   while(m<=r)temp[temp_pos++]=a[m++];
   int i;
   for( i=0;i<size;i++){
       a[r]=temp[r];
       r--;
   }
}

void mergeSort(struct student a[],struct student temp[],int l,int r){
   if(l<r){
       int mid=l+(r-l)/2;
       mergeSort(a,temp,l,mid);
       mergeSort(a,temp,mid+1,r);
       merge(a,temp,l,mid+1,r);
   }
}

int main(){
   int n,i;
   printf("Enter number of students : ");
   scanf("%d",&n);
  
   struct student st[n];
  
   printf("Enter data of %d students\n",n);
   for(i=0;i<n;i++){
       printf("Enter detail of student %d\n",i+1);
       printf("Enter name : ");
       scanf(" %[^\n]%*c", st[i].name);
       printf("Enter Id : ");
       scanf("%d",&st[i].ID);
       printf("Enter GPA : ");
       scanf("%lf",&st[i].GPA);
   }
  
   printf("Sorting on the basis of ID using Bubble Sort (O(n^2)).............\n");
   bubbleSort(st,n);
   printData(st,n);
  
   printf("\n\nSorting on the basis of GPA using Merge Sort (O(nlogn)).............\n");
   struct student temp[n];
   mergeSort(st,temp,0,n-1);
   printData(st,n);
  
   return 0;
}

//sample output

Enter number of students : 5 Enter data of 5 students Enter detail of student 1 Enter name : abc Enter Id : 5 Enter GPA : 4.1 Enter detail of student 2 Enter name : xyz Enter Id : 2 Enter GPA : 3.2 Enter detail of student 3 Enter nane : que Enter Id : 1 Enter GPA : 5.0 Enter detail of student 4 Enter name : def Enter Id : 3 Enter GPA : 2.1 Enter detail of student 5 Enter name : jkl Enter Id : 4 Enter GPA : 4.9 Sorting on the basis of ID using Bubble Sort con^2>>.... Student 1 Name : qwe ID : 1 GPA : 5.00 Student 2 Nane : xyz ID : 2 GPA : 3.20 Student 3 Name : def ID : 3 GPA : 2.10 Student 4 Name : jkl ID : 4 GPA : 4.90 Student 5 Name : abc ID: 5 GPA : 4.10

Sorting on the basis of GPA using Merge Sort (Ocnlogn>>.. Student 1 Name : def ID : 3 GPA : 2.10 ------------------------------- Student 2 Name : xyz ID : 2 GPA : 3.20 ------ ---------- --- Student 3 Name : abc ID : 5 GPA : 4.10 ------------------- Student 4 Name : jkl ID : 4 GPA : 4.90 - --------- --------- Student 5 Name : que ID : 1 GPA : 5.00 ------ ---------- ------ - Process exited after 83.62 seconds with return value o Press any key to continue - - -

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