# Rearrange an array to minimize sum of product of consecutive pair elements in C++

C++Server Side ProgrammingProgramming

We are given a positive integer type array, let's say, arr[] of any given size. The task is to rearrange an array in such a manner that when we multiply an element with its alternative element and then add all the resultant elements then it should return the minimum sum.

## Let us see various input output scenarios for this −

Input − int arr[] = {2, 5, 1, 7, 5, 0, 1, 0}

Output − Rearrangement of an array to minimize sum i.e. 7 of product of consecutive pair elements is: 7 0 5 0 5 1 2 1

Explanation − we are given an integer array of size 8. Now, we will rearrange the array i.e. 7 0 5 0 5 1 2 1. We will check whether its returning minimum sum i.e. 7 * 0 + 5 * 0 + 5 * 1 + 2 * 1 = 0 + 0 + 5 + 2 = 7.

Input − int arr[] = {1, 3, 7, 2, 4, 3}

Output − Rearrangement of an array to minimize sum i.e. 24 of product of consecutive pair elements is: 7 1 4 2 3 3

Explanation − we are given an integer array of size 6. Now, we will rearrange the array i.e. 7 1 4 2 3 3. We will check whether its returning minimum sum i.e. 7 * 1 + 4 * 2 + 3 * 3 = 7 + 8 + 9 = 24.

## Approach used in the below program is as follows

• Input an array of integer type elements and calculate the size of an array.

• Sort an array using the sort method of C++ STL by passing array and size of an array to the sort function.

• Declare an integer variable and set it with the call to the function Rearrange_min_sum(arr, size)

• Inside the function Rearrange_min_sum(arr, size)

• Create a variable, let's say, ‘even’ and ‘odd’ type of type vector which stores integer variables.

• Declare a variable as temp and total and initialise it with 0.

• Start loop FOR from i to 0 till i less than size. Inside the loop, check IF i is less than size/2 then push arr[i] to odd vector ELSE, push arr[i] to even vector

• Call the sort method by passing even.begin(), even.end() and greater<int>().

• Start loop FOR from i to 0 till i less than even.size(). Inside the loop, set arr[temp++] to even[j], arr[temp++] to odd[j] and total to total + even[j] * odd[j]

• Print the result.

## Example

#include <bits/stdc++.h>
using namespace std;
int Rearrange_min_sum(int arr[], int size){
vector<int> even, odd;
int temp = 0;
int total = 0;
for(int i = 0; i < size; i++){
if (i < size/2){
odd.push_back(arr[i]);
}
else{
even.push_back(arr[i]);
}
}
sort(even.begin(), even.end(), greater<int>());
for(int j = 0; j < even.size(); j++){
arr[temp++] = even[j];
arr[temp++] = odd[j];
total += even[j] * odd[j];
}
}
int main(){
int arr[] = { 2, 5, 1, 7, 5, 0, 1, 0};
int size = sizeof(arr)/sizeof(arr);
//sort an array
sort(arr, arr + size);
//call function
int total = Rearrange_min_sum(arr, size);
cout<<"Rearrangement of an array to minimize sum i.e. "<<total<<" of product of consecutive pair elements is: ";
for(int i = 0; i < size; i++){
cout << arr[i] << " ";
}
return 0;
}

## Output

If we run the above code it will generate the following Output

Rearrangement of an array to minimize sum i.e. 7 of product of consecutive pair elements is: 7 0 5 0 5 1 2 1