Maximum sum of i*arr[i] among all rotations of a given array
Last Updated :
28 Jun, 2025
Given an integer array arr[] of size n, find the maximum value of the expression i * arr[i] (for all i from 0 to n-1) after rotating the array any number of times.
Examples:
Input: arr[] = [8, 3, 1, 2]
Output: 29
Explanation: Out of all the possible configurations by rotating the elements: arr[] = [3, 1, 2, 8] here (3*0) + (1*1) + (2*2) + (8*3) sum is maximum i.e. 29.
Input: arr[] = [1, 2, 3]
Output: 8
Explanation: Out of all the possible configurations by rotating the elements: arr[] = [1, 2, 3] here (1*0) + (2*1) + (3*2) sum is maximum i.e. 8.
[Naive Approach] Using Nested Loops - O(n ^ 2) Time and O(1) Space
The idea is to find the sum of all the elements of the array for each possible rotation, and store the maximum of them as answer.
- To do so, use nested loops, where the outer loop marks the starting point of the array, and the inner loop iterates through each of the element, and store the sum of i * arr[i].
- Check if the maximum sum is greater than the current sum then update the maximum sum.
C++
#include <iostream>
#include <vector>
#include <limits.h>
using namespace std;
int maxSum(vector<int> &arr) {
int n = arr.size();
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 sum = 0;
// Compute sum of all values
for (int j = 0; j < n; j++) {
// compute the rotated index
int index = (i + j) % n;
sum += j * arr[index];
}
// Update the maximum value
res = max(res, sum);
}
return res;
}
int main() {
vector<int> arr = {8, 3, 1, 2};
cout << maxSum(arr);
return 0;
}
class GfG {
static int maxSum(int[] arr) {
int n = arr.length;
// 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 sum = 0;
// Compute sum of all values
for (int j = 0; j < n; j++) {
// compute the rotated index
int index = (i + j) % n;
sum += j * arr[index];
}
// Update the maximum value
res = Math.max(res, sum);
}
return res;
}
public static void main(String[] args) {
int[] arr = {8, 3, 1, 2};
System.out.println(maxSum(arr));
}
}
def maxSum(arr):
n = len(arr)
# Initialize result
res = -float('inf')
# Consider rotation beginning with i
# for all possible values of i.
for i in range(n):
# Initialize sum of current rotation
sum = 0
# Compute sum of all values
for j in range(n):
# compute the rotated index
index = (i + j) % n
sum += j * arr[index]
# Update the maximum value
res = max(res, sum)
return res
if __name__ == "__main__":
arr = [8, 3, 1, 2]
print(maxSum(arr))
using System;
class GfG {
static int maxSum(int[] arr) {
int n = arr.Length;
// 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 sum = 0;
// Compute sum of all values
for (int j = 0; j < n; j++) {
// compute the rotated index
int index = (i + j) % n;
sum += j * arr[index];
}
// Update the maximum value
res = Math.Max(res, sum);
}
return res;
}
static void Main() {
int[] arr = {8, 3, 1, 2};
Console.WriteLine(maxSum(arr));
}
}
function maxSum(arr) {
let n = arr.length;
// Initialize result
let res = -Infinity;
// Consider rotation beginning with i
// for all possible values of i.
for (let i = 0; i < n; i++) {
// Initialize sum of current rotation
let sum = 0;
// Compute sum of all values
for (let j = 0; j < n; j++) {
// compute the rotated index
let index = (i + j) % n;
sum += j * arr[index];
}
// Update the maximum value
res = Math.max(res, sum);
}
return res;
}
// Driver Code
let arr = [8, 3, 1, 2];
console.log(maxSum(arr));
[Expected Approach] Using Mathematics - O(n) Time and O(1) Space
The main idea is to Instead of recalculating the weighted sum from scratch for each rotation, it first computes the total sum of the array and the initial rotation value. Then, for each subsequent rotation, it derives the new value using a formula based on the previous value and the total sum. and find the maximum value of the sum across all rotations of the array.
We can calculate the value of the next rotation using the value of the current one. Let curSum be the total sum of all elements in the array, and currVal be the value of the current rotation. Then, the value of the next rotation can be derived from currVal by adjusting it based on the shift in element positions.
finding NextVal Using currVal:
We can compute the value of the ith rotation using the value of the (i-1)th rotation:
Nextval = currVal - (curSum - arr[i-1]) + arr[i-1] * (n-1)
C++
#include <iostream>
#include <vector>
using namespace std;
int maxSum(vector<int> &arr) {
int n = arr.size();
// Compute sum of all array elements
int curSum = 0;
for (int i = 0; i < n; i++)
curSum += arr[i];
// Compute sum of i*arr[i] for initial
// configuration.
int currVal = 0;
for (int i = 0; i < n; i++)
currVal += i * arr[i] ;
// Initialize result
int res = currVal ;
// Compute values for other iterations
for (int i = 1; i < n; i++) {
// Compute next value using previous
int nextVal = currVal - (curSum - arr[i - 1]) +
arr[i - 1] * (n - 1);
// Update current value
currVal = nextVal;
// Update result if required
res = max(res, nextVal);
}
return res;
}
int main() {
vector<int> arr = {8, 3, 1, 2};
cout << maxSum(arr);
return 0;
}
class GfG {
static int maxSum(int[] arr) {
int n = arr.length;
// Compute sum of all array elements
int curSum = 0;
for (int i = 0; i < n; i++)
curSum += arr[i];
// Compute sum of i*arr[i] for initial
// configuration.
int currVal = 0;
for (int i = 0; i < n; i++)
currVal += i * arr[i];
// Initialize result
int res = currVal;
// Compute values for other iterations
for (int i = 1; i < n; i++) {
// Compute next value using previous
int nextVal = currVal - (curSum - arr[i - 1]) +
arr[i - 1] * (n - 1);
// Update current value
currVal = nextVal;
// Update result if required
res = Math.max(res, nextVal);
}
return res;
}
public static void main(String[] args) {
int[] arr = {8, 3, 1, 2};
System.out.println(maxSum(arr));
}
}
def maxSum(arr):
n = len(arr)
# Compute sum of all array elements
curSum = 0
for i in range(n):
curSum += arr[i]
# Compute sum of i*arr[i] for initial
# configuration.
currVal = 0
for i in range(n):
currVal += i * arr[i]
# Initialize result
res = currVal
# Compute values for other iterations
for i in range(1, n):
# Compute next value using previous
nextVal = currVal - (curSum - arr[i - 1]) \
+ arr[i - 1] * (n - 1)
# Update current value
currVal = nextVal
# Update result if required
res = max(res, nextVal)
return res
if __name__ == "__main__":
arr = [8, 3, 1, 2]
print(maxSum(arr))
using System;
class GfG {
static int maxSum(int[] arr) {
int n = arr.Length;
// Compute sum of all array elements
int curSum = 0;
for (int i = 0; i < n; i++)
curSum += arr[i];
// Compute sum of i*arr[i] for initial
// configuration.
int currVal = 0;
for (int i = 0; i < n; i++)
currVal += i * arr[i];
// Initialize result
int res = currVal;
// Compute values for other iterations
for (int i = 1; i < n; i++) {
// Compute next value using previous
int nextVal = currVal - (curSum - arr[i - 1]) +
arr[i - 1] * (n - 1);
// Update current value
currVal = nextVal;
// Update result if required
res = Math.Max(res, nextVal);
}
return res;
}
static void Main() {
int[] arr = {8, 3, 1, 2};
Console.WriteLine(maxSum(arr));
}
}
function maxSum(arr) {
let n = arr.length;
// Compute sum of all array elements
let curSum = 0;
for (let i = 0; i < n; i++)
curSum += arr[i];
// Compute sum of i*arr[i] for initial
// configuration.
let currVal = 0;
for (let i = 0; i < n; i++)
currVal += i * arr[i];
// Initialize result
let res = currVal;
// Compute values for other iterations
for (let i = 1; i < n; i++) {
// Compute next value using previous
let nextVal = currVal - (curSum - arr[i - 1]) +
arr[i - 1] * (n - 1);
// Update current value
currVal = nextVal;
// Update result if required
res = Math.max(res, nextVal);
}
return res;
}
// Driver Code
let arr = [8, 3, 1, 2];
console.log(maxSum(arr));
[Alternate Approach] Using Pivot - O(n) Time and O(1) Space
Note: This approach works only for sorted or rotated sorted arrays.
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.
- 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
C++
#include <iostream>
#include <vector>
using namespace std;
// Function to find pivot
int findPivot(vector<int> &arr) {
int n = arr.size();
for(int i = 0; i < n; i++) {
if(arr[i] > arr[(i + 1) % n])
return i;
}
return 0;
}
// Function to find maximum sum rotation
int maxSum(vector<int> &arr) {
int n = arr.size();
// initialize result
int res = 0;
int pivot = findPivot(arr);
// difference in pivot and index of
// last element of array
int diff = n - 1 - pivot;
// compute the sum
for(int i = 0; i < n; i++) {
res = res + ((i + diff) % n) * arr[i];
}
return res;
}
int main() {
vector<int> arr = {8, 3, 1, 2};
cout << maxSum(arr);
return 0;
}
class GfG {
// Function to find pivot
static int findPivot(int[] arr) {
int n = arr.length;
for (int i = 0; i < n; i++) {
if (arr[i] > arr[(i + 1) % n])
return i;
}
return 0;
}
// Function to find maximum sum rotation
static int maxSum(int[] arr) {
int n = arr.length;
// initialize result
int res = 0;
int pivot = findPivot(arr);
// difference in pivot and index of
// last element of array
int diff = n - 1 - pivot;
// compute the sum
for (int i = 0; i < n; i++) {
res = res + ((i + diff) % n) * arr[i];
}
return res;
}
public static void main(String[] args) {
int[] arr = {8, 3, 1, 2};
System.out.println(maxSum(arr));
}
}
def findPivot(arr):
n = len(arr)
for i in range(n):
if arr[i] > arr[(i + 1) % n]:
return i
return 0
# Function to find maximum sum rotation
def maxSum(arr):
n = len(arr)
# initialize result
res = 0
pivot = findPivot(arr)
# difference in pivot and index of
# last element of array
diff = n - 1 - pivot
# compute the sum
for i in range(n):
res = res + ((i + diff) % n) * arr[i]
return res
if __name__ == "__main__":
arr = [8, 3, 1, 2]
print(maxSum(arr))
using System;
class GfG {
// Function to find pivot
static int findPivot(int[] arr) {
int n = arr.Length;
for (int i = 0; i < n; i++) {
if (arr[i] > arr[(i + 1) % n])
return i;
}
return 0;
}
// Function to find maximum sum rotation
static int maxSum(int[] arr) {
int n = arr.Length;
// initialize result
int res = 0;
int pivot = findPivot(arr);
// difference in pivot and index of
// last element of array
int diff = n - 1 - pivot;
// compute the sum
for (int i = 0; i < n; i++) {
res = res + ((i + diff) % n) * arr[i];
}
return res;
}
static void Main() {
int[] arr = {8, 3, 1, 2};
Console.WriteLine(maxSum(arr));
}
}
function findPivot(arr) {
let n = arr.length;
for (let i = 0; i < n; i++) {
if (arr[i] > arr[(i + 1) % n])
return i;
}
return 0;
}
// Function to find maximum sum rotation
function maxSum(arr) {
let n = arr.length;
// initialize result
let res = 0;
let pivot = findPivot(arr);
// difference in pivot and index of
// last element of array
let diff = n - 1 - pivot;
// compute the sum
for (let i = 0; i < n; i++) {
res = res + ((i + diff) % n) * arr[i];
}
return res;
}
// Driver Code
let arr = [8, 3, 1, 2];
console.log(maxSum(arr));
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