Question

Subsequence with Minimum Average Difference
Given a sequence of n (>10) integers, not all the elements of S are same, write a pseudo-code and the subsequent C program, to compute the arithmetic average of the sequences called k-th sequence of
(here k < 1/2), denoted as Sk, formed from the sequence S, by removing the first k elements of S and the last k-elements of the sequence Your program should print the pair subsequences whose
absolute difference of the averages is minimum.
For eg, consider the sequences: < 2,4 6 7,8 9,5,7,6, 1>
S1: <4,6,7,8,9,5,7,6> and AS1 = 6.5
S2: <6,7,8,9,5,7> and AS2 - 7
S3: <7,8,9,5> and AS3 - 7.25
O CSE1001:Problem Solvi 17:29
S4: <8,9> and AS4 = 8.5
Absolute difference between the averages of the sequences are:
S1 S2 - 05
here to search

Answer 1

Answer:

Subsequence with Minimum Average Difference

Given a sequence of n (>10) integers, not all the elements of S are same, write a pseudo-code and the subsequent C program, to compute the arithmetic average of the sequences called k-th sequence of

(here k < 1/2), denoted as Sk, formed from the sequence S, by removing the first k elements of S and the last k-elements of the sequence Your program should print the pair subsequences whose

absolute difference of the averages is minimum.

For eg, consider the sequences: < 2,4 6 7,8 9,5,7,6, 1>

S1: <4,6,7,8,9,5,7,6> and AS1 = 6.5

S2: <6,7,8,9,5,7> and AS2 - 7

S3: <7,8,9,5> and AS3 - 7.25

O CSE1001:Problem Solvi 17:29

S4: <8,9> and AS4 = 8.5

Absolute difference between the averages of the sequences are:

S1 S2 - 05

here to search

Answer 2

Subsequence with Minimum Average Difference

Explanation:

#include <stdio.h>

#include <math.h>

int main() {

   int n;

   scanf("%d", &n);

   int arri[n];

   for (int i=0; i<n; i++){

       scanf("%d", &arri[i]);

   }

   printf("Array= ");

   for (int i=0; i<n;i++){

       printf("%d ",arri[i]);

   }

   int mid=(n+1)/2;

   float AM[mid];

   for (int j; j<mid; j++){

       int sum=0;

       for (int i=j; i<mid;i++){

           if ((n%2==1) && i==mid-1) sum+=arri[i];

           else sum+=arri[i]+arri[n-i-1];

       }

       printf("\nfor S%d sum=%d", j, sum);

       float am= sum/(float)(n-2*j);

       printf("\tam=%f",am);

       AM[j]=am;

   }

   float diff=9999999;

   int X,Y;

   for (int x=0; x<mid;x++){

       float temp=0;

       for (int y=x+1; y<mid;y++){

           temp=fabs(AM[y]-AM[x]);

           if (temp<diff) {

             diff=temp;

             X=x;

             Y=y;

           }

       }

   }

   printf("\nMin difference=%f, between sequences S%d, S%d", diff, X, Y);

   

   return 0;

}