Minimise sum of products of positions of each element with element at (i+K)%N

Given an array Arr[] of integers of size N, The task is to find the minimum value of the given function when applied on the array.

S(K) =  1 * Arr[(0 + K)%N] + 2 * Arr[(1 + K)%N] + 3 * Arr[(2 + K)%N] + . . . . . . . + N * Arr[((N-1) + K)%N]
        = ∑[(i+1)∗Arr[(i+K)%N]

where K is a whole number.

Examples:

Input: Arr[] = {2, 1, 4, 3}
Output: 20
Explanation: S(0) = 28, S(1) = 26, S(2) = 20, S(3) =26, S(4) = S(0), S(5) = S(1). . . . .
Thus there are only four possible values of the function and minimum is 20.

Input: Arr[] = {-3, 2, 1}
Output: -5
Explanation: S(0) = 4, S(1) = -5, S(2) = 1, S(3) = S(0), S(4) = S(1), S(5) = S(2) . . . . .
Thus there are only three possible values of the function and minimum is -5.

Naive Approach: The basic approach to solve the problem is to calculate the value of the function for all values of K in range [0, N-1].

Follow the steps mentioned below to solve the problem:

• It can be observed that if the Arr has length N then the given function can only have N distinct values.
• Find all the possible values of the function S(k) separately from k = 0 to N – 1 and print the minimum value among them.

Below is the implementation of the naive approach.

C++

// C++ Program to find minimum value of given function
#include <bits/stdc++.h>
using namespace std;

// Function to find value
// of given function at k
int fnValueAtk(vector<int> Arr, int k)
{
  int sum = 0;
  int N = Arr.size();
  for (int i = 0; i < N; i++) {
    sum += (i + 1) * Arr[(i + k) % N];
  }
  return sum;
}

// Function to print minimum value
// of given function
void minValFn(vector<int> Arr)
{
  int N = Arr.size();
  int min_val = INT_MAX;
  for (int k = 0; k < N; k++) {
    int fn_At_k = fnValueAtk(Arr, k);
    min_val = min(min_val, fn_At_k);
  }
  cout << min_val;
}

int main()
{

  vector<int> Arr = { 2, 1, 4, 3 };
  minValFn(Arr);
  return 0;
}

// This code is contributed by hrithikgarg03188.

Java

// Java Program to find minimum value of given function

import java.util.*;

class GFG {

    // Function to print minimum value
    // of given function
    static void minValFn(int[] Arr)
    {
        int N = Arr.length;
        int min_val = Integer.MAX_VALUE;
        for (int k = 0; k < N; k++) {
            int fn_At_k = fnValueAtk(Arr, k);
            min_val = Math.min(min_val, fn_At_k);
        }
        System.out.println(min_val);
    }

    // Function to find value
    // of given function at k
    static int fnValueAtk(int[] Arr, int k)
    {
        int sum = 0;
        int N = Arr.length;
        for (int i = 0; i < N; i++) {
            sum += (i + 1) * Arr[(i + k) % N];
        }
        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] Arr = { 2, 1, 4, 3 };
        minValFn(Arr);
    }
}

Python3

# Python3 program to find the minimum value of given function
import sys

# function to find the value
# of given function at k
def fnValueAtk(Arr, k):
    sums = 0
    N = len(Arr)
    for i in range(N):
        sums += (i + 1) * Arr[(i + k) % N]
    return sums

# function to print minimum value of given function
def minValFn(Arr):
    N = len(Arr)
    min_val = sys.maxsize
    for k in range(N):
        fn_At_k = fnValueAtk(Arr, k)
        min_val = min(min_val, fn_At_k)

    print(min_val)

# Driver Code
Arr = [2, 1, 4, 3]
minValFn(Arr)

# This code is contributed by phasing17

C#

// C# Program to find minimum value of given function
using System;

public class GFG{

  // Function to print minimum value
  // of given function
  static void minValFn(int[] Arr)
  {
    int N = Arr.Length;
    int min_val = Int32.MaxValue;
    for (int k = 0; k < N; k++) {
      int fn_At_k = fnValueAtk(Arr, k);
      min_val = Math.Min(min_val, fn_At_k);
    }
    Console.WriteLine(min_val);
  }

  // Function to find value
  // of given function at k
  static int fnValueAtk(int[] Arr, int k)
  {
    int sum = 0;
    int N = Arr.Length;
    for (int i = 0; i < N; i++) {
      sum += (i + 1) * Arr[(i + k) % N];
    }
    return sum;
  }

  // Driver code
  static public void Main (){

    int[] Arr = { 2, 1, 4, 3 };
    minValFn(Arr);
  }
}

// This code is contributed by hrithikgarg03188.

Javascript

// JavaScript program to find the
// minimum value of the given function

// function to find the value of
// given function at k
function fnValueAtk(Arr, k)
{
  var sum = 0;
  var N = Arr.length;
  for (var i = 0; i < N; i++)
  {
    sum += (i + 1) * Arr[(i + k) % N];
  }
  return sum;
}

// function to print the minimum value
// of given function
function minValFn(Arr)
{
  var N = Arr.length;
  var min_val = Number.MAX_SAFE_INTEGER;
  for (var k = 0; k < N; k++)
  {
    var fn_At_k = fnValueAtk(Arr, k);
    min_val = Math.min(min_val, fn_At_k);
  }
  document.write(min_val);
}

// Driver Code
var Arr = [2, 1, 4, 3];
minValFn(Arr);

// This code is contributed by phasing17

Output

20

Time Complexity: O(N²).
Auxiliary Space: O(1)

Efficient approach: The efficient approach to solve the problem is based on the following relations among S(k) and S(k-1):

S(k) = S(k – 1) –  Sum + Arr[(k-1)%N] * N

Refer to the illustration below for a better understanding.

Illustration:

Suppose the array is {2, 1, 4, 3}

Now, to find the value of the function given in the problem:

=  1 * Arr[(0 + k)%N] + 2 * Arr[(1 + k)%N] + 3 * Arr[(2 + k)%N] + . . . . . . . + N * Arr[((N-1) + k)%N]

• Now, for k = 0;
  S(0) = 1* Arr[0] + 2* Arr[1] + 3* Arr[2] + 4* Arr[3]
         = 1*2 + 2*1 + 3*4 + 4*3
         = 28
• similarly, for k = 1;
  S(1) = 1* Arr[1] + 2* Arr[2] + 3* Arr[3] + 4* Arr[0]
         = 1*1 + 2*4 + 3*3 + 4*2
         = 26
• similarly, for k = 2;
  S(2) = 20
• similarly, for k = 3;
  s(3) = 26
• Now, S(4) = S(0), S(5) = S(1). . . . and so on.

Hence the minimum value of function is S(2) = 20;

Now, it can be solved using derived equation:  S(k) = S(k – 1) –  Sum + Arr[(k-1)%N] * N

Firstly, store the value of S(0) (k = 0) in some variable by traversing the array and then use the above mentioned equation to find the values of function for k = 1, 2, 3 and so on. 
Later compare the value of S(k) with S(0) and update it with the minimum one. Follow the below steps:

• Now, the value of function when k = 0 S(0) is 28, and total sum is 10,  then to find the value for k = 1
  using equation,  
  for k = 1: S(1)  = 28 – 10 + Arr[(1 – 1) % 4] * 4
                         = 28 – 10 + 8
                         = 26 (which is less than S(0) hence update the minimum value with S(1))
• For k = 2: S(2) = 26 – 10 + Arr[(2 – 1) % 4]* 4
                         = 26 – 10 + 4
                         = 20 (which is less than S(1) hence update the minimum value with S(2))
• For k = 3: S(3)  = 20 – 10 + Arr[(3 – 1) % 4]* 4
                          = 20 – 10  + 16
                          = 26 (which is not less than S(2) hence, do not update the value with S(3))

Hence, S(2) is the minimum value of function.

Follow the steps mentioned below to implement the above observation:

• Declare a variable ‘min_val’ to store the minimum value of the given function.
• Find S(0) in a single traversal of the array and store it in ‘min_val’.
• Then find all the possible values of function S(k) for k = 1 to N – 1 using the above equation.
  • If this value is smaller than ‘min_val’ then update ‘min_val’.
• The final value of min_val is the required answer.

Below is the implementation of the above approach:

C++

// C++ Program to find minimum value
// Of given function

#include <bits/stdc++.h>
using namespace std;

void minValFn(vector<int> Arr)
{
    int N = Arr.size();
    int min_val;

    // To store sum of all elements
    int sum = 0;

    // Finding S(0)
    int s0 = 0;
    for (int i = 0; i < N; i++) {
        s0 += Arr[i] * (i + 1);
        sum += Arr[i];
    }

    min_val = s0;

    // To store s(k-1)
    int sk_min_1 = s0;
    for (int k = 1; k < N; k++) {

        // Using formula
        int sk = sk_min_1 - sum + Arr[(k - 1) % N] * N;
        min_val = min(sk, min_val);
        sk_min_1 = sk;
    }
    cout << min_val << "\n";
}

// Driver code
int main()
{
    vector<int> Arr = { 2, 1, 4, 3 };
    minValFn(Arr);
}

// This code is contributed by Taranpreet

Java

// Java Program to find minimum value
// Of given function

import java.io.*;

public class GFG {
    static void minValFn(int[] Arr)
    {
        int N = Arr.length;
        int min_val;

        // To store sum of all elements
        int sum = 0;

        // Finding S(0)
        int s0 = 0;
        for (int i = 0; i < N; i++) {
            s0 += Arr[i] * (i + 1);
            sum += Arr[i];
        }

        min_val = s0;

        // To store s(k-1)
        int sk_min_1 = s0;
        for (int k = 1; k < N; k++) {

            // Using formula
            int sk = sk_min_1 - sum
                     + Arr[(k - 1) % N] * N;
            min_val = Math.min(sk, min_val);
            sk_min_1 = sk;
        }
        System.out.println(min_val);
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] Arr = { 2, 1, 4, 3 };
        minValFn(Arr);
    }
}

Python3

# Python3 program to find the minimum value of
# the given function

# function to find the minimum value
def minValFn(Arr):
    N = len(Arr)

    # to store sum of all elements
    sums = 0

    # finding S(0)
    s0 = 0
    for i in range(N):
        s0 += Arr[i] * (i + 1)
        sums += Arr[i]

    min_val = s0

    # To store s(k - 1)
    sk_min_1 = s0
    for k in range(1, N):
        # using formula for minimum value
        sk = sk_min_1 - sums + Arr[(k - 1) % N] * N
        min_val = min(sk, min_val)
        sk_min_1 = sk

    print(min_val)

# Driver code
Arr = [2, 1, 4, 3]
minValFn(Arr)

# This code is contributed by phasing17

C#

// C# Program to find minimum value
// Of given function

using System;

public class GFG {
  static void minValFn(int[] Arr)
  {
    int N = Arr.Length;
    int min_val;

    // To store sum of all elements
    int sum = 0;

    // Finding S(0)
    int s0 = 0;
    for (int i = 0; i < N; i++) {
      s0 += Arr[i] * (i + 1);
      sum += Arr[i];
    }

    min_val = s0;

    // To store s(k-1)
    int sk_min_1 = s0;
    for (int k = 1; k < N; k++) {

      // Using formula
      int sk = sk_min_1 - sum
        + Arr[(k - 1) % N] * N;
      min_val = Math.Min(sk, min_val);
      sk_min_1 = sk;
    }
    Console.Write(min_val);
  }

  // Driver code
  public static void Main(string[] args)
  {
    int[] Arr = { 2, 1, 4, 3 };
    minValFn(Arr);
  }
}

// This code is contributed by ukasp.

JavaScript

<script>
    // JavaScript Program to find minimum value
    // Of given function


    const minValFn = (Arr) => {
        let N = Arr.length;
        let min_val;

        // To store sum of all elements
        let sum = 0;

        // Finding S(0)
        let s0 = 0;
        for (let i = 0; i < N; i++) {
            s0 += Arr[i] * (i + 1);
            sum += Arr[i];
        }

        min_val = s0;

        // To store s(k-1)
        let sk_min_1 = s0;
        for (let k = 1; k < N; k++) {

            // Using formula
            let sk = sk_min_1 - sum + Arr[(k - 1) % N] * N;
            min_val = Math.min(sk, min_val);
            sk_min_1 = sk;
        }
        document.write(min_val);
    }

    // Driver code

    let Arr = [2, 1, 4, 3];
    minValFn(Arr);

    // This code is contributed by rakeshsahni

</script>

Output

20

Time Complexity: O(N)
Auxiliary Space: O(1)