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

04 Jul, 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:Rotate the array for all values from 0 to n.Calculate the sum for each rotations.Check if the maximum sum is greater than the current sum then update the maximum sum.
Rotate the array for all values from 0 to n.Calculate the sum for each rotations.Check if the maximum sum is greater than the current sum then update the maximum sum.
Rotate the array for all values from 0 to n.
Calculate the sum for each rotations.
Check if the maximum sum is greater than the current sum then update the maximum sum.
Implementation:
C++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){    int arr[] = {8, 3, 1, 2};    int n = sizeof(arr)/sizeof(arr[0]);    cout << maxSum(arr, n) << endl;    return 0;}
// A Naive Java program to find// maximum sum rotationimport java.util.*;import java.io.*; class GFG { // Returns maximum value of i*arr[i]static int maxSum(int arr[], int n){// Initialize resultint 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 codepublic 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
# A Naive Python3 program to find# maximum sum rotationimport 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 codearr = [8, 3, 1, 2]n = len(arr) print(maxSum(arr, n)) # This code is contributed by# Smitha Dinesh Semwal
// A Naive C# program to find// maximum sum rotationusing 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// 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?>
<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>
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Complexity Analysis: Time Complexity : O(n2)Auxiliary Space : O(1)
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]. Multiplier of arr[i-1] changes from 0 to n-1, i.e., arr[i-1] * (n-1) is added to current value.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.
Multiplier of arr[i-1] changes from 0 to n-1, i.e., arr[i-1] * (n-1) is added to current value.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.
Multiplier of arr[i-1] changes from 0 to n-1, i.e., arr[i-1] * (n-1) is added to current value.
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++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){    int arr[] = {8, 3, 1, 2};    int n = sizeof(arr)/sizeof(arr[0]);    cout << maxSum(arr, n) << endl;    return 0;}
// 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
# 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 codearr = [8, 3, 1, 2]n = len(arr) print(maxSum(arr, n)) # This code is contributed by# Smitha Dinesh Semwal
// 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// 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?>
<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>
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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)
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: 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.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
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.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
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.
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
Java
Python3
C#
PHP
Javascript
// C++ program to find maximum sum of all// rotation of i*arr[i] using pivot.#include <iostream>using namespace std; // fun declarationint maxSum(int arr[], int n);int findPivot(int arr[], int n); // function definitionint 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 pivotint findPivot(int arr[], int n){    int i;    for(i = 0; i < n; i++)    {        if(arr[i] > arr[(i + 1) % n])            return i;    }} // Driver codeint main(void){         // rotated input array    int arr[] = {8, 3, 1, 2};    int n = sizeof(arr) / sizeof(int);    int max = maxSum(arr, n);    cout << max;    return 0;} // This code is contributed by Shubhamsingh10
// C program to find maximum sum of all// rotation of i*arr[i] using pivot.#include<stdio.h> // fun declarationint maxSum(int arr[], int n);int findPivot(int arr[], int n); // function definitionint 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 pivotint findPivot(int arr[], int n){    int i;    for(i = 0; i < n; i++)    {        if(arr[i] > arr[(i + 1) % n])            return i;    }} // Driver codeint main(void){         // rotated input array    int arr[] = {8, 3, 1, 2};    int n = sizeof(arr) / sizeof(int);    int max = maxSum(arr, n);    printf("%d", max);    return 0;}
// 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 definitionstatic 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 pivotstatic 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 codepublic static void main(String args[]){    // rotated input array    int arr[] = {8, 3, 1, 2};    int n = arr.length;    int max = maxSum(arr,n);    System.out.println(max);     }}
# Python3 program to find maximum sum of# all rotation of i*arr[i] using pivot. # function definitiondef 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 pivotdef findPivot(arr, n) :    for i in range(n) :         if(arr[i] > arr[(i + 1) % n]) :            return i; # Driver codeif __name__ == "__main__" :     # rotated input array    arr = [8, 3, 1, 2]    n= len(arr)         max= maxSum(arr, n)    print(max) # This code is contributed by Ryuga
// C# program to find maximum sum// of all rotation of i*arr[i] using pivot.using System; class GFG{ // function definitionpublic 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 pivotpublic 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 codepublic static void Main(string[] args){    // rotated input array    int[] arr = new int[] {8, 3, 1, 2};    int n = arr.Length;    int max = maxSum(arr,n);    Console.WriteLine(max);}} // This code is contributed by Shrikant13
<?php// PHP program to find maximum sum// of all rotation of i*arr[i] using pivot. // function definitionfunction 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 $sum; } // function to find pivotfunction findPivot($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, 3, 1, 2);$n = sizeof($arr);$max = maxSum($arr, $n);echo $max; // This code is contributed// by Akanksha Rai(Abby_akku)?>
<script> // js program to find maximum sum// of all rotation of i*arr[i] using pivot. // function definitionfunction 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 pivotfunction 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 arraylet arr = [8, 3, 1, 2];let n = arr.length;let ma = maxSum(arr,n);document.write(ma); // This code is contributed by mohit kumar 29.</script>
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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 to so the Auxiliary space is O(1)
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 to so the Auxiliary space is O(1)
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