Question

C programming for the below question..

In this program we are going to test the performance of two searching codes.
(a) Write a randomarray(n, max) function that returns a pointer to an array of size n integers,
filled with random values between 0 and max.
(b) Write a median(n, arr) function that returns the median of an integer array, without ordering
them. The median value means that half the numbers are smaller or equal and half are
bigger or equal. If the length n is even, choose the n/2 highest value. In this implementation
of the median, the return value must be a number within the sequence.
i. Starting from the first element determine if it is the median.
ii. If not advance to the next element and check, repeat the process until the median is
found.
iii. It is possible to skip elements to make it more efficient.

Answer 1

Thanks for the question.

Below is the code you will be needing Let me know if you have any doubts or if you need anything to change.

Thank You !!

===========================================================================

#include<stdio.h>
#include<stdlib.h>

int* randomarray(int n, int max);
int median(int n, int arr[]);

int* randomarray(int n, int max){
  
   int* pointer = (int*)malloc(n*sizeof(int));
  
   int i;
  
   for(i=0; i<n; i++){
      
       *(pointer+i) = rand()%max;
   }
  
   return pointer;
}

int median(int n, int arr[]){
  
   int i,j;
  
   // logic to find median when n is odd
   for(i=0; i<n; i++){
       int bigger_count=0;
       int smaller_count=0;
      
       for(j=0; j<n; j++){
           if(i==j)continue;
           if(arr[j]<=arr[i])bigger_count+=1;
           else if (arr[j]>=arr[i])smaller_count+=1;
       }
      
       if(bigger_count==smaller_count && n%2==1)return arr[i];
   }  
   // logic to find median when n is even
   int two_median[2]={-1,-1};
   int index=0;
  
   for(i=0; i<n; i++){
       int bigger_count=0;
       int smaller_count=0;
      
       for(j=0; j<n; j++){
           if(i==j)continue;
           if(arr[j]<=arr[i])bigger_count+=1;
           else if (arr[j]>=arr[i])smaller_count+=1;
       }
// when count of numbers bigger than the current number is same as the count of numbers smaller than the current

// number
       if(bigger_count-smaller_count==1 || bigger_count-smaller_count==-1){
           two_median[index]=arr[i];
           index=index+1;
       }
       if(index==2)break;
   }
   if(two_median[0]<two_median[1])   return two_median[1];
   else return two_median[0];
}

int main(){
   int n = 10;
   int max=100;
   int *p = randomarray(n,max);
   int median_ = median(n,p);
   printf("Median = %d\n\n",median_);
  
   int i=0;
   for(i=0; i<n; i++){
       printf("%d ",*(p+i));
   }
  
   free(p);
  
}

=====================================================================

rain.cpp median.cpp #include<stdio.h 1 #include<stdlib. h> int* randomarray (int n, int max) int median(int n, int arr [] ); CAUsers User Documents\median.exe 6 7E int* randomarray ( int n, int max){ Median 62 int* pointer = (int*)malloc (n*sizeof (int)); 9 41 67 34 69 24 78 58 62 64 10 int i 11. Process exited after 0.07344 seconds with return value 0 Press any key to continue . . . 13 F for (i-0; i<n; it+){ *(pointer+i) rand()%max; 15 17 return pointer; 18 int median(int n, int arr[]){ 23 24 int i,j // logic to find median when n is odd for (i-0; i<n; i++){ int bigger_count-0; int smaller_count-0; 30 Col: 24 Line: 13 Sel: 0 Lines: 79 Length: 1431