Find minimum value of K to maximise sum of elements on indices that are multiples of K

Given an array arr[] of N integers, the task is to find the minimum value of K such that the sum of elements on indices that are multiples of K is the maximum possible.

Example:

Input: arr[] = {-3, 4}
Output: 2
Explanation: For the given array, it the value of K = 1, then the multiples of K are {1, 2} which sums up to arr[1] + arr[2] = -3 + 4 = 1. For K = 2, the valid multiple of K is 2 and hence the sum is arr[2] = 4, which is the maximum possible. Hence, K = 2 is a valid answer. 

Input: arr[] = {-1, -2, -3}
Output: 2

Approach: The given problem can be solved by a similar to that of the Sieve of Eratosthenes. The idea is to calculate the sum for all possible values of K in the range [1, N] by iterating over each multiple of K as done while marking the non-prime elements in the Sieve. The value of K giving the maximum sum is the required answer.

Below is the implementation of the above approach:

C++

// C++ Program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find minimum K such that
// the sum of elements on indices that
// are multiples of K is maximum possible
int maxSum(int arr[], int N)
{
    // Stores the maximum sum and
    // respective K value
    int maxSum = INT_MIN, ans = -1;

    // Loop to iterate over all
    // value of K in range [1, N]
    for (int K = 1; K <= N; K++) {
        int sum = 0;

        // Iterating over all
        // multiples of K
        for (int i = K; i <= N; i += K) {
            sum += arr[i - 1];
        }

        // Update Maximum Sum
        if (sum > maxSum) {
            maxSum = sum;
            ans = K;
        }
    }

    // Return Answer
    return ans;
}

// Driver Code
int main()
{
    int arr[] = { -1, -2, -3 };
    int N = sizeof(arr) / sizeof(int);

    cout << maxSum(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find minimum K such that
// the sum of elements on indices that
// are multiples of K is maximum possible
static int maxSum(int arr[], int N)
{

    // Stores the maximum sum and
    // respective K value
    int maxSum = Integer.MIN_VALUE, ans = -1;

    // Loop to iterate over all
    // value of K in range [1, N]
    for(int K = 1; K <= N; K++)
    {
        int sum = 0;

        // Iterating over all
        // multiples of K
        for(int i = K; i <= N; i += K)
        {
            sum += arr[i - 1];
        }

        // Update Maximum Sum
        if (sum > maxSum)
        {
            maxSum = sum;
            ans = K;
        }
    }

    // Return Answer
    return ans;
}

// Driver Code
public static void main(String args[])
{
    int arr[] = { -1, -2, -3 };
    int N = arr.length;

    System.out.println(maxSum(arr, N));
}
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach
import sys

# Function to find minimum K such that
# the sum of elements on indices that
# are multiples of K is maximum possible
def maxSum(arr, N):

    # Stores the maximum sum and
    # respective K value
    maxSum = -sys.maxsize;
    ans = -1;

    # Loop to iterate over all
    # value of K in range [1, N]
    for K in range(1,N+1):
        sum = 0;

        # Iterating over all
        # multiples of K
        for i in range(K, N + 1,K):
            sum += arr[i - 1];

        # Update Maximum Sum
        if (sum > maxSum):
            maxSum = sum;
            ans = K;

    # Return Answer
    return ans;

# Driver Code
if __name__ == '__main__':
    arr = [ -1, -2, -3 ];
    N = len(arr);

    print(maxSum(arr, N));

# This code is contributed by gauravrajput1

C#

// C# program for the above approach
using System;

public class GFG{

  // Function to find minimum K such that
  // the sum of elements on indices that
  // are multiples of K is maximum possible
  static int maxSum(int []arr, int N)
  {

    // Stores the maximum sum and
    // respective K value
    int maxSum = int.MinValue, ans = -1;

    // Loop to iterate over all
    // value of K in range [1, N]
    for(int K = 1; K <= N; K++)
    {
      int sum = 0;

      // Iterating over all
      // multiples of K
      for(int i = K; i <= N; i += K)
      {
        sum += arr[i - 1];
      }

      // Update Maximum Sum
      if (sum > maxSum)
      {
        maxSum = sum;
        ans = K;
      }
    }

    // Return Answer
    return ans;
  }

  // Driver Code
  public static void Main(String []args)
  {
    int []arr = { -1, -2, -3 };
    int N = arr.Length;

    Console.WriteLine(maxSum(arr, N));
  }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
      // JavaScript code for the above approach


      // Function to find minimum K such that
      // the sum of elements on indices that
      // are multiples of K is maximum possible
      function maxSum(arr, N) {
          // Stores the maximum sum and
          // respective K value
          let maxSum = -999999;
          let ans = -1;

          // Loop to iterate over all
          // value of K in range [1, N]
          for (let K = 1; K <= N; K++) {
              let sum = 0;

              // Iterating over all
              // multiples of K
              for (let i = K; i <= N; i += K) {
                  sum = sum + arr[i - 1];
              }

              // Update Maximum Sum
              if (sum > maxSum) {
                  maxSum = sum;
                  ans = K;
              }
          }

          // Return Answer
          return ans;
      }

      // Driver Code

      let arr = [-1, -2, -3];
      let N = arr.length;

      document.write(maxSum(arr, N));


// This code is contributed by Potta Lokesh
  </script>

Output

2

Time Complexity: O(N*log N)
Auxiliary space: O(1)

Another Approach:

• First, we calculate the prefix sum of the given array. Let’s say prefixSum[i] stores the sum of the first i elements of the array.
• Then, we iterate over all possible values of K in the range [1, N] and calculate the sum of elements on indices that are multiples of K. This can be done in constant time by using prefixSum array. Let’s say this sum is sumK.
• We update the maximum sum and the corresponding value of K whenever we find a new maximum sum.
• Finally, we return the value of K that gave us the maximum sum.
  

Below is the implementation of the above approach:

C++

#include <iostream>
#include <vector>
using namespace std;

int maxSum(vector<int>& arr) {
    int n = arr.size();
    vector<int> prefixSum(n+1, 0); // prefixSum[i] stores sum of first i elements of arr
    for(int i=1; i<=n; i++) {
        prefixSum[i] = prefixSum[i-1] + arr[i-1];
    }

    int maxSum = -999999999, ans = -1;
    for(int K=1; K<=n; K++) {
        int sumK = 0;
        for(int i=K; i<=n; i+=K) {
            sumK += prefixSum[i] - prefixSum[i-K];
        }
        if(sumK > maxSum) {
            maxSum = sumK;
            ans = K;
        }
    }

    return ans;
}

int main() {
    vector<int> arr = {-1, -2, -3};
    cout << maxSum(arr) << endl;
    return 0;
}

Output

2

Time Complexity: O(n*logn)

Auxiliary Space: O(1)