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Maximise product of each array element with their indices by rearrangement

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  • Difficulty Level : Easy
  • Last Updated : 17 Aug, 2022
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Given an Array of N integers, the task is to maximize the value of the product of each array element with their corresponding indices by rearranging the array.

Examples:

Input: N = 4, arr[] = { 3, 5, 6, 1 }
Output: 31
Explanation: If we arrange arr[] as { 1, 3, 5, 6 }. Sum of arr[i]*i is 1*0 + 3*1 + 5*2 + 6*3 = 31, which is maximum

Input: N = 2, arr[] = { 19, 20 }
Output: 20

Maximise product of each array element with their indices by rearrangement using Sorting

The idea is based on the fact that the largest value should be scaled maximum and the smallest value should be scaled minimum. So we multiply the minimum value of i with the minimum value of arr[i]. So, sort the given array in increasing order and compute the sum of arr[i]*i.

Follow the steps mentioned below to implement the idea:

  • Sort the array.
  • Create a variable sum to store the final answer
  • Traverse The array, while traversing the array increases the value of the sum by arr[I]*i.
  • Return sum.

Below is the implementation of the above approach:

C++




// CPP program to find the maximum value
// of i*arr[i]
#include <bits/stdc++.h>
using namespace std;
 
int maxSum(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
 
    // Finding the sum of arr[i]*i
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += (arr[i] * i);
 
    return sum;
}
 
// Driven Program
int main()
{
    int arr[] = { 3, 5, 6, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxSum(arr, n) << endl;
    return 0;
}


Java




// Java program to find the
// maximum value of i*arr[i]
import java.util.*;
 
class GFG {
 
    static int maxSum(int arr[], int n)
    {
        // Sort the array
        Arrays.sort(arr);
 
        // Finding the sum of arr[i]*i
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += (arr[i] * i);
 
        return sum;
    }
 
    // Driven Program
    public static void main(String[] args)
    {
        int arr[] = { 3, 5, 6, 1 };
        int n = arr.length;
 
        System.out.println(maxSum(arr, n));
    }
}
// This code is contributed by Prerna Saini


Python3




# Python program to find the
# maximum value of i*arr[i]
 
 
def maxSum(arr, n):
 
    #  Sort the array
    arr.sort()
 
    # Finding the sum of
    # arr[i]*i
    sum = 0
    for i in range(n):
        sum += arr[i] * i
 
    return sum
 
 
# Driver Program
arr = [3, 5, 6, 1]
n = len(arr)
print(maxSum(arr, n))
 
# This code is contributed
# by Shrikant13


C#




// C# program to find the
// maximum value of i*arr[i]
using System;
 
class GFG {
 
    // Function to find the
    // maximum value of i*arr[i]
    static int maxSum(int[] arr, int n)
    {
 
        // Sort the array
        Array.Sort(arr);
 
        // Finding the sum of arr[i]*i
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += (arr[i] * i);
 
        return sum;
    }
 
    // Driver code
    static public void Main()
    {
        int[] arr = { 3, 5, 6, 1 };
        int n = arr.Length;
 
        Console.WriteLine(maxSum(arr, n));
    }
}
 
// This code is contributed by Ajit.


PHP




<?php
// PHP program to find the
// maximum value of i*arr[i]
 
// function returns the
// maximum value of i*arr[i]
function maxSum($arr, $n)
{
    // Sort the array
    sort($arr);
     
    // Finding the sum
    // of arr[i]*i
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        $sum += ($arr[$i] * $i);
     
    return $sum;
}
 
// Driver Code
$arr = array( 3, 5, 6, 1 );
$n = count($arr);
 
echo maxSum($arr, $n);
 
// This code is contributed by anuj_67.
?>


Javascript




<script>
// JavaScript program to find the
// maximum value of i*arr[i]
 
function maxSum(arr, n)
    {   
    // Sort the array
    arr.sort();
 
    // Finding the sum of arr[i]*i
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += (arr[i] * i);
 
    return sum;
    }
  
// Driver Code
 
    let arr = [ 3, 5, 6, 1 ];
    let n = arr.length;
 
    document.write(maxSum(arr, n));
         
</script>


Output

31

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


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