Maximum sum of i*arr[i] among all rotations of a given array

Last Updated : 14 Dec, 2022

Given an array arr[] of n integers, find the maximum that maximizes the sum of the value of i*arr[i] where i varies from 0 to n-1.

Examples:

Input: arr[] = {8, 3, 1, 2}
Output: 29
Explanation: Lets look at all the rotations,
{8, 3, 1, 2} = 8*0 + 3*1 + 1*2 + 2*3 = 11
{3, 1, 2, 8} = 3*0 + 1*1 + 2*2 + 8*3 = 29
{1, 2, 8, 3} = 1*0 + 2*1 + 8*2 + 3*3 = 27
{2, 8, 3, 1} = 2*0 + 8*1 + 3*2 + 1*3 = 17

Input: arr[] = {3, 2, 1}
Output: 7
Explanation: Lets look at all the rotations,
{3, 2, 1} = 3*0 + 2*1 + 1*2 = 4
{2, 1, 3} = 2*0 + 1*1 + 3*2 = 7
{1, 3, 2} = 1*0 + 3*1 + 2*2 = 7

Method 1: This method discusses the Naive Solution which takes O(n²) amount of time.
The solution involves finding the sum of all the elements of the array in each rotation and then deciding the maximum summation value.

Approach:A simple solution is to try all possible rotations. Compute sum of i*arr[i] for every rotation and return maximum sum.
Algorithm:
1. Rotate the array for all values from 0 to n.
2. Calculate the sum for each rotations.
3. Check if the maximum sum is greater than the current sum then update the maximum sum.

Implementation:

C++

// A Naive C++ program to find maximum sum rotation
#include<bits/stdc++.h>
 
using namespace std;
 
// Returns maximum value of i*arr[i]
int maxSum(int arr[], int n)
{
   // Initialize result
   int res = INT_MIN;
 
   // Consider rotation beginning with i
   // for all possible values of i.
   for (int i=0; i<n; i++)
   {
 
     // Initialize sum of current rotation
     int curr_sum = 0;
 
     // Compute sum of all values. We don't
     // actually rotate the array, instead of that we compute the
     // sum by finding indexes when arr[i] is
     // first element
     for (int j=0; j<n; j++)
     {
         int index = (i+j)%n;
         curr_sum += j*arr[index];
     }
 
     // Update result if required
     res = max(res, curr_sum);
   }
 
   return res;
}
 
// Driver code
int main()
{
    int arr[] = {8, 3, 1, 2};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << maxSum(arr, n) << endl;
    return 0;
}

Java

// A Naive Java program to find
// maximum sum rotation
import java.util.*;
import java.io.*;
 
class GFG {
 
// Returns maximum value of i*arr[i]
static int maxSum(int arr[], int n)
{
// Initialize result
int res = Integer.MIN_VALUE;
 
// Consider rotation beginning with i
// for all possible values of i.
for (int i = 0; i < n; i++)
{
 
    // Initialize sum of current rotation
    int curr_sum = 0;
 
    // Compute sum of all values. We don't
    // actually rotation the array, but compute
    // sum by finding indexes when arr[i] is
    // first element
    for (int j = 0; j < n; j++)
    {
        int index = (i + j) % n;
        curr_sum += j * arr[index];
    }
 
    // Update result if required
    res = Math.max(res, curr_sum);
}
 
return res;
}
 
// Driver code
public static void main(String args[])
{
        int arr[] = {8, 3, 1, 2};
        int n = arr.length;
        System.out.println(maxSum(arr, n));
}
 
     
}
 
// This code is contributed by Sahil_Bansall

Python3

# A Naive Python3 program to find
# maximum sum rotation
import sys
 
# Returns maximum value of i * arr[i]
def maxSum(arr, n):
 
    # Initialize result
    res = -sys.maxsize
 
    # Consider rotation beginning with i
    # for all possible values of i.
    for i in range(0, n):
 
 
        # Initialize sum of current rotation
        curr_sum = 0
     
        # Compute sum of all values. We don't
        # actually rotation the array, but 
        # compute sum by finding indexes when 
        # arr[i] is first element
        for j in range(0, n):
         
            index = int((i + j)% n) 
            curr_sum += j * arr[index] 
     
 
        # Update result if required
        res = max(res, curr_sum)
    return res 
 
# Driver code
arr = [8, 3, 1, 2] 
n = len(arr)
 
print(maxSum(arr, n))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// A Naive C# program to find
// maximum sum rotation
using System;
 
class GFG {
 
    // Returns maximum value of i*arr[i]
    static int maxSum(int[] arr, int n)
    {
        // Initialize result
        int res = int.MinValue;
 
        // Consider rotation beginning with i
        // for all possible values of i.
        for (int i = 0; i < n; i++) {
 
            // Initialize sum of current rotation
            int curr_sum = 0;
 
            // Compute sum of all values. We don't
            // actually rotate the array, instead we compute
            // sum by finding indexes when arr[i] is the
            // first element
            for (int j = 0; j < n; j++) 
            {
                int index = (i + j) % n;
                curr_sum += j * arr[index];
            }
 
            // Update result if required
            res = Math.Max(res, curr_sum);
        }
 
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 8, 3, 1, 2 };
        int n = arr.Length;
        Console.WriteLine(maxSum(arr, n));
    }
}
 
// This code is contributed by vt_m.

PHP

<?php
// A Naive PHP program to 
// find maximum sum rotation
 
// Returns maximum value
// of i*arr[i]
function maxSum($arr, $n)
{
// Initialize result
$res = PHP_INT_MIN;
 
// Consider rotation beginning 
// with i for all possible 
// values of i.
for ($i = 0; $i < $n; $i++)
{
 
    // Initialize sum of
    // current rotation
    $curr_sum = 0;
 
    // Compute sum of all values.
    // We don't actually rotate 
    // the array, but compute sum
    // by finding indexes when 
    // arr[i] is first element
    for ($j = 0; $j < $n; $j++)
    {
        $index = ($i + $j) % $n;
        $curr_sum += $j * $arr[$index];
    }
 
    // Update result if required
    $res = max($res, $curr_sum);
}
return $res;
}
 
// Driver code
$arr = array(8, 3, 1, 2);
$n = sizeof($arr);
echo maxSum($arr, $n), "\n";
 
// This code is contributed by ajit
?>

Javascript

<script>
// A Naive javascript program to find
// maximum sum rotation    // Returns maximum value of i*arr[i]
    function maxSum(arr , n) {
        // Initialize result
        var res = Number.MIN_VALUE;
 
        // Consider rotation beginning with i
        // for all possible values of i.
        for (i = 0; i < n; i++) {
 
            // Initialize sum of current rotation
            var curr_sum = 0;
 
            // Compute sum of all values. We don't
            // actually rotation the array, but compute
            // sum by finding indexes when arr[i] is
            // first element
            for (j = 0; j < n; j++) {
                var index = (i + j) % n;
                curr_sum += j * arr[index];
            }
 
            // Update result if required
            res = Math.max(res, curr_sum);
        }
 
        return res;
    }
 
    // Driver code
     
        var arr = [ 8, 3, 1, 2 ];
        var n = arr.length;
        document.write(maxSum(arr, n));
 
// This code contributed by umadevi9616
</script>

Output

Complexity Analysis:
- Time Complexity : O(n²)
- Auxiliary Space : O(1)

Method 2: This method discusses the efficient solution which solves the problem in O(n) time. In the naive solution, the values were calculated for every rotation. So if that can be done in constant time then the complexity will decrease.

Approach: The basic approach is to calculate the sum of new rotation from the previous rotations. This brings up a similarity where only the multipliers of first and last element change drastically and the multiplier of every other element increases or decreases by 1. So in this way, the sum of next rotation can be calculated from the sum of present rotation.
Algorithm:
The idea is to compute the value of a rotation using values of previous rotation. When an array is rotated by one, following changes happen in sum of i*arr[i].
1. Multiplier of arr[i-1] changes from 0 to n-1, i.e., arr[i-1] * (n-1) is added to current value.
2. Multipliers of other terms is decremented by 1. i.e., (cum_sum – arr[i-1]) is subtracted from current value where cum_sum is sum of all numbers.

next_val = curr_val - (cum_sum - arr[i-1]) + arr[i-1] * (n-1);

next_val = Value of ?i*arr[i] after one rotation.
curr_val = Current value of ?i*arr[i] 
cum_sum = Sum of all array elements, i.e., ?arr[i].

Lets take example {1, 2, 3}. Current value is 1*0+2*1+3*2
= 8. Shifting it by one will make it {2, 3, 1} and next value
will be 8 - (6 - 1) + 1*2 = 5 which is same as 2*0 + 3*1 + 1*2

Implementation:

C++

// An efficient C++ program to compute
// maximum sum of i*arr[i]
#include<bits/stdc++.h>
 
using namespace std;
 
int maxSum(int arr[], int n)
{
    // Compute sum of all array elements
    int cum_sum = 0;
    for (int i=0; i<n; i++)
        cum_sum += arr[i];
 
    // Compute sum of i*arr[i] for initial
    // configuration.
    int curr_val = 0;
    for (int i=0; i<n; i++)
        curr_val += i*arr[i];
 
    // Initialize result
    int res = curr_val;
 
    // Compute values for other iterations
    for (int i=1; i<n; i++)
    {
        // Compute next value using previous 
        // value in O(1) time
        int next_val = curr_val - (cum_sum - arr[i-1])
                       + arr[i-1] * (n-1);
 
        // Update current value
        curr_val = next_val;
 
        // Update result if required
        res = max(res, next_val);
    }
 
    return res;
}
 
// Driver code
int main()
{
    int arr[] = {8, 3, 1, 2};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << maxSum(arr, n) << endl;
    return 0;
}

Java

// An efficient Java program to compute
// maximum sum of i*arr[i]
import java.io.*;
 
class GFG {
     
    static int maxSum(int arr[], int n)
    {
        // Compute sum of all array elements
        int cum_sum = 0;
        for (int i = 0; i < n; i++)
            cum_sum += arr[i];
 
        // Compute sum of i*arr[i] for 
        // initial configuration.
        int curr_val = 0;
        for (int i = 0; i < n; i++)
            curr_val += i * arr[i];
 
        // Initialize result
        int res = curr_val;
 
        // Compute values for other iterations
        for (int i = 1; i < n; i++)
        {
            // Compute next value using previous
            // value in O(1) time
            int next_val = curr_val - (cum_sum -
                          arr[i-1]) + arr[i-1] *
                          (n-1);
 
            // Update current value
            curr_val = next_val;
 
            // Update result if required
            res = Math.max(res, next_val);
        }
 
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {8, 3, 1, 2};
        int n = arr.length;
        System.out.println(maxSum(arr, n));
    }
}
// This code is contributed by Prerna Saini

Python3

# An efficient Python3 program to
# compute maximum sum of i * arr[i]
 
def maxSum(arr, n):
 
    # Compute sum of all array elements
    cum_sum = 0
     
    for i in range(0, n):
        cum_sum += arr[i] 
 
    # Compute sum of i * arr[i] for 
    # initial configuration.
    curr_val = 0
     
    for i in range(0, n):
        curr_val += i * arr[i] 
 
    # Initialize result
    res = curr_val 
 
    # Compute values for other iterations
    for i in range(1, n):
     
        # Compute next value using previous
        # value in O(1) time
        next_val = (curr_val - (cum_sum - arr[i-1]) +
                                    arr[i-1] * (n-1))
 
        # Update current value
        curr_val = next_val 
 
        # Update result if required
        res = max(res, next_val)
     
    return res 
 
 
# Driver code
arr = [8, 3, 1, 2] 
n = len(arr)
 
print(maxSum(arr, n))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// An efficient C# program to compute
// maximum sum of i*arr[i]
using System;
 
class GFG {
     
    static int maxSum(int []arr, int n)
    {
         
        // Compute sum of all array elements
        int cum_sum = 0;
        for (int i = 0; i < n; i++)
            cum_sum += arr[i];
 
        // Compute sum of i*arr[i] for 
        // initial configuration.
        int curr_val = 0;
        for (int i = 0; i < n; i++)
            curr_val += i * arr[i];
 
        // Initialize result
        int res = curr_val;
 
        // Compute values for other iterations
        for (int i = 1; i < n; i++)
        {
             
            // Compute next value using previous
            // value in O(1) time
            int next_val = curr_val - (cum_sum -
                      arr[i - 1]) + arr[i - 1] *
                                        (n - 1);
 
            // Update current value
            curr_val = next_val;
 
            // Update result if required
            res = Math.Max(res, next_val);
        }
 
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = {8, 3, 1, 2};
        int n = arr.Length;
        Console.Write(maxSum(arr, n));
    }
}
 
// This code is contributed by nitin mittal

PHP

<?php
// An efficient PHP program to 
// compute maximum sum of i*arr[i]
 
function maxSum($arr, $n)
{
    // Compute sum of all
    // array elements
    $cum_sum = 0;
    for ($i = 0; $i < $n; $i++)
        $cum_sum += $arr[$i];
 
    // Compute sum of i*arr[i] 
    // for initial configuration.
    $curr_val = 0;
    for ($i = 0; $i < $n; $i++)
        $curr_val += $i * $arr[$i];
 
    // Initialize result
    $res = $curr_val;
 
    // Compute values for
    // other iterations
    for ($i = 1; $i < $n; $i++)
    {
        // Compute next value using 
        // previous value in O(1) time
        $next_val = $curr_val - 
                   ($cum_sum - $arr[$i - 1]) + 
                    $arr[$i - 1] * ($n - 1);
 
        // Update current value
        $curr_val = $next_val;
 
        // Update result if required
        $res = max($res, $next_val);
    }
 
    return $res;
}
 
// Driver code
$arr = array(8, 3, 1, 2);
$n = sizeof($arr);
echo maxSum($arr, $n);
 
// This code is contributed by ajit
?>

Javascript

<script>
// An efficient JavaScript program to compute
// maximum sum of i*arr[i]
 
function maxSum(arr, n)
{
    // Compute sum of all array elements
    let cum_sum = 0;
    for (let i=0; i<n; i++)
        cum_sum += arr[i];
 
    // Compute sum of i*arr[i] for initial
    // configuration.
    let curr_val = 0;
    for (let i=0; i<n; i++)
        curr_val += i*arr[i];
 
    // Initialize result
    let res = curr_val;
 
    // Compute values for other iterations
    for (let i=1; i<n; i++)
    {
        // Compute next value using previous
        // value in O(1) time
        let next_val = curr_val - (cum_sum - arr[i-1])
                    + arr[i-1] * (n-1);
 
        // Update current value
        curr_val = next_val;
 
        // Update result if required
        res = Math.max(res, next_val);
    }
 
    return res;
}
 
// Driver code
    let arr = [8, 3, 1, 2];
    let n = arr.length;
    document.write(maxSum(arr, n) + "<br>");
 
 
 
 
 
// This code is contributed by Surbhi Tyagi.
</script>

Output

Complexity analysis:
- Time Complexity: O(n).
  Since one loop is needed from 0 to n to check all rotations and the sum of the present rotation is calculated from the previous rotations in O(1) time).
- Auxiliary Space: O(1).
  As no extra space is required to so the space complexity will be O(1)

Method 3: The method discusses the solution using pivot in O(n) time. The pivot method can only be used in the case of a sorted or a rotated sorted array. For example: {1, 2, 3, 4} or {2, 3, 4, 1}, {3, 4, 1, 2} etc.

Approach: Let’s assume the case of a sorted array. As we know for an array the maximum sum will be when the array is sorted in ascending order. In case of a sorted rotated array, we can rotate the array to make it in ascending order. So, in this case, the pivot element is needed to be found following which the maximum sum can be calculated.
Algorithm:
1. Find the pivot of the array: if arr[i] > arr[(i+1)%n] then it is the pivot element. (i+1)%n is used to check for the last and first element.
2. After getting pivot the sum can be calculated by finding the difference with the pivot which will be the multiplier and multiply it with the current element while calculating the sum

Implementations:

C++

// C++ program to find maximum sum of all 
// rotation of i*arr[i] using pivot. 
#include <iostream>
using namespace std; 
 
// fun declaration 
int maxSum(int arr[], int n); 
int findPivot(int arr[], int n); 
 
// function definition 
int maxSum(int arr[], int n) 
{ 
    int sum = 0; 
    int i; 
    int pivot = findPivot(arr, n); 
 
    // difference in pivot and index of 
    // last element of array 
    int diff = n - 1 - pivot; 
    for(i = 0; i < n; i++) 
    { 
        sum = sum + ((i + diff) % n) * arr[i]; 
    } 
    return sum; 
} 
 
// function to find pivot 
int findPivot(int arr[], int n) 
{ 
    int i; 
    for(i = 0; i < n; i++) 
    { 
        if(arr[i] > arr[(i + 1) % n]) 
            return i; 
    } 
} 
 
// Driver code 
int main(void) 
{ 
     
    // rotated input array 
    int arr[] = {8, 13, 1, 2}; 
    int n = sizeof(arr) / sizeof(int); 
    int max = maxSum(arr, n); 
    cout << max; 
    return 0; 
} 
 
// This code is contributed by Shubhamsingh10 

C

// C program to find maximum sum of all 
// rotation of i*arr[i] using pivot.
#include<stdio.h>
 
// fun declaration
int maxSum(int arr[], int n); 
int findPivot(int arr[], int n);
 
// function definition
int maxSum(int arr[], int n) 
{
    int sum = 0;
    int i;
    int pivot = findPivot(arr, n);
 
    // difference in pivot and index of 
    // last element of array
    int diff = n - 1 - pivot; 
    for(i = 0; i < n; i++)
    {
        sum= sum + ((i + diff) % n) * arr[i];
    }
    return sum;
}
 
// function to find pivot
int findPivot(int arr[], int n)
{
    int i;
    for(i = 0; i < n; i++)
    {
        if(arr[i] > arr[(i + 1) % n])
            return i;
    }
}
 
// Driver code
int main(void)
{
     
    // rotated input array
    int arr[] = {8, 13, 1, 2}; 
    int n = sizeof(arr) / sizeof(int);
    int max = maxSum(arr, n); 
    printf("%d", max);
    return 0; 
}

Java

// Java program to find maximum sum 
// of all rotation of i*arr[i] using pivot.
 
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
 
// function definition 
static int maxSum(int arr[], int n) 
{
    int sum = 0;
    int i;
    int pivot = findPivot(arr, n);
 
    // difference in pivot and index of
    // last element of array
    int diff = n - 1 - pivot; 
    for(i = 0; i < n; i++)
    { 
        sum= sum + ((i + diff) % n) * arr[i];
    }
    return sum;
}
 
// function to find pivot
static int findPivot(int arr[], int n)
{
    int i;
    for(i = 0; i < n; i++)
    {
        if(arr[i] > arr[(i + 1) % n])
            return i;
    }
    return 0;
}
 
// Driver code
public static void main(String args[])
{
    // rotated input array
    int arr[] = {8, 13, 1, 2}; 
    int n = arr.length;
    int max = maxSum(arr,n); 
    System.out.println(max);
     
}
}

Python3

# Python3 program to find maximum sum of
# all rotation of i*arr[i] using pivot.
 
# function definition
 
 
def maxSum(arr, n):
 
    sum = 0
    pivot = findPivot(arr, n)
 
    # difference in pivot and index
    # of last element of array
    diff = n - 1 - pivot
    for i in range(n):
        sum = sum + ((i + diff) % n) * arr[i]
 
    return sum
 
# function to find pivot
 
 
def findPivot(arr, n):
    for i in range(n):
 
        if(arr[i] > arr[(i + 1) % n]):
            return i
 
 
# Driver code
if __name__ == "__main__":
 
    # rotated input array
    arr = [8, 13, 1, 2]
    n = len(arr)
 
    max = maxSum(arr, n)
    print(max)
 
# This code is contributed by Ryuga

C#

// C# program to find maximum sum 
// of all rotation of i*arr[i] using pivot. 
using System;
 
class GFG
{
 
// function definition 
public static int maxSum(int[] arr, int n)
{
    int sum = 0;
    int i;
    int pivot = findPivot(arr,n);
     
    // difference in pivot and index of 
    // last element of array 
    int diff = n - 1 - pivot;
    for (i = 0;i < n;i++)
    {
        sum = sum + ((i + diff) % n) * arr[i];
    }
     
    return sum;
 
}
 
// function to find pivot 
public static int findPivot(int[] arr, int n)
{
    int i;
    for (i = 0; i < n; i++)
    {
        if (arr[i] > arr[(i + 1) % n])
        {
            return i;
        }
    }
    return 0;
    }
 
// Driver code 
public static void Main(string[] args)
{
    // rotated input array 
    int[] arr = new int[] {8, 13, 1, 2};
    int n = arr.Length;
    int max = maxSum(arr,n);
    Console.WriteLine(max);
}
}
 
// This code is contributed by Shrikant13

PHP

<?php
// PHP program to find maximum sum 
// of all rotation of i*arr[i] using pivot.
 
// function definition 
function maxSum($arr, $n) 
{
$sum = 0;
$pivot = findPivot($arr, $n);
 
// difference in pivot and index of
// last element of array
$diff = $n - 1 - $pivot; 
for($i = 0; $i < $n; $i++)
{
    $sum = $sum + (($i + $diff) % 
            $n) * $arr[$i];
}
 
return $sum;
 
}
 
// function to find pivot
function findPivot($arr, $n)
{
 
    for($i = 0; $i < $n; $i++)
    {
        if($arr[$i] > $arr[($i + 1) % $n])
        return $i;
    }
    return 0;
}
 
// Driver code
 
// rotated input array
$arr = array(8, 13, 1, 2); 
$n = sizeof($arr);
$max = maxSum($arr, $n); 
echo $max;
 
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

Javascript

<script>
 
// js program to find maximum sum 
// of all rotation of i*arr[i] using pivot. 
 
// function definition 
function maxSum(arr, n)
{
    let sum = 0;
    let i;
    let pivot = findPivot(arr,n);
     
    // difference in pivot and index of 
    // last element of array 
    let diff = n - 1 - pivot;
    for (i = 0;i < n;i++)
    {
        sum = sum + ((i + diff) % n) * arr[i];
    }
     
    return sum;
 
}
 
// function to find pivot 
function findPivot(arr, n)
{
    let i;
    for (i = 0; i < n; i++)
    {
        if (arr[i] > arr[(i + 1) % n])
        {
            return i;
        }
    }
    return 0;
}
 
// Driver code 
 
// rotated input array 
let arr = [8, 13, 1, 2];
let n = arr.length;
let ma = maxSum(arr,n);
document.write(ma);
 
// This code is contributed by mohit kumar 29.
</script>

Output

Complexity analysis:
- Time Complexity: O(n)
  As only one loop was needed to traverse from 0 to n to find the pivot. To find the sum another loop was needed, so the complexity remains O(n).
- Auxiliary Space: O(1).
  We do not require extra space so the Auxiliary space is O(1)

Suggest improvement

Print array after it is right rotated K times

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Maximum sum of i*arr[i] among all rotations of a given array

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