Given an array arr that contains n integers, the following operation can be performed on it any number of times (possibly zero):
Choose any index 1 ≤ i ≤ n and swap arr[i]. Each element of the array can be swapped at most once during the whole process.
The strength of an index is defined as (arr[i] + 1) using 1-based indexing. Find the maximum possible sum of the strength of all indices after optimal swaps. Mathematically, maximize the following:
Complete the function getMaximumSumOfStrengths in the editor below.
getMaximumSumOfStrengths has the following parameter:
int arr[n]: the initial array
Returns
long int: the maximum sum of strengths
arr = [1, 9, 7, 3, 2] return = 66
It is optimal to swap (arr[2], arr[3]). The final array is arr[] = [1, 9, 3, 7, 2]. The sum of strengths (1*1 + 2*9 + 3*3 + 4*7 + 5*2) = 66, which is the maximum possible.
arr = [2, 1, 4, 3] return = 30
It is optimal to swap (arr[0], arr[1]) and (arr[2], arr[3]). The final array is arr[] = [1, 2, 3, 4]. The sum of strengths (1*1 + 2*2 + 3*3 + 4*4) = 30, which is the maximum possible.
arr = [1, 2, 5] return = 20
No swaps are needed as the array is already in optimal order. The sum of strengths (1*1 + 2*2 + 3*5) = 20, which is the maximum possible.
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public long getMaximumSumOfStrengths(int[] arr) {
// write your code here
}