We are given with an array of integers of size N and a number k. The array consists of integers in random order. The task is to find the maximum difference between the group of k-elements and rest of the array. The array will be divided into two parts. The first part is a group of k-elements taken out and the second part is the rest of the elements of the array. We have to select k elements such that the difference between the sum of elements in both groups is maximum.
If k is smaller (<= half of array size) then smallest k elements will have least sum and rest N-k elements will have largest sum. So the maximum difference is − (sum of rest N-k elements) - (sum of smallest k-elements).
If k is larger (> half of array size) then largest k elements will have the greatest sum and rest N-k elements will have least sum. So the maximum difference is (sum of largest kelements) - (sum of rest N-k elements).
Input
Arr[] = { 2,5,6,1,3,2,1,4 }. k=3Output − Maximum difference between the group of k-elements and rest of the array − 16
Explanation − Here k is smaller so the least 3 numbers would have the smallest sum.
Least 3 numbers − 1,1,2 sum=4
Rest N-k = 5 numbers: 2,3,4,5,6 sum=20
Maximum difference : 20-4 = 16
Input
Arr[] = { 2,2,3,4,8,3,4,4,8,7 }. k=6Output − Maximum difference between the group of k-elements and rest of the array − 25
Explanation − Here k is larger so the highest 6 numbers would have the largest sum.
Highest 6 numbers − 8,8,7,4,4,4, sum=35
Rest N-k = 4 numbers − 2,2,3,3 sum=10
Maximum difference − 35-10=
Approach used in the below program is as follows
Declare an array of integers which contains in random order.( Arr[] )
Create a variable to store the size of the array. (N)
The function maxKDiff( int Arr[],int n, int k) is used to calculate the maximum difference (maxD) between first and last indexes of an element in an array.
Calculate the sum of whole array and store in arrsum.
First thing is to calculate the sum of least k elements. Using for loop ( i=0;i<k)
If k is smaller then the smallest k elements would have least sum −
In D1 store the abs((sum of whole array) - (2*sum of least k elements)). Twice because array sum also has these elements.
k is larger then the largest k elements would have highest sum −
In D2 store the abs((sum of whole array) - (2*sum of highest k elements)). Twice because array sum also has these elements.
Compare D1 with D2 and store maximum value in maxD.
Return maxD as the result.
Example
#include <stdio.h>
#include <math.h>
// function for finding maximum group difference of array
int maxKDiff (int arr[], int n, int k){
// sum of array
int arrsum = 0;
int i;
for(i=0;i<n;i++)
arrsum+=arr[i];
//sum of smallest k
int sumk=0;
for(i=0;i<k;i++)
sumk+=arr[i];
// difference for k-smallest
int D1 = abs(arrsum - 2*sumk);
//sum of largest k elements
sumk=0;
int j=0;
for(i=n-1;j<4;i--){
sumk+=arr[i]; j++;
}
// difference for k-largest
int D2 = abs(arrsum - 2*sumk);
int maxD=D1>=D2?D1:D2;
// return maximum difference value
return maxD;
}
// driver program
int main(){
int arr[] ={ 2,3,2,10,7,12,8};
int n = 7;
int k = 3;
sort(arr,n); // to sort array in ascending order
printf("Maximum difference between the group of k-elements and rest of the array : %d" , maxKDiff(arr,n,k));
return 0;
}Output
If we run the above code it will generate the following output −
Maximum difference between the group of k-elements and rest of the array : 30