Salesforce OA || 24 sept

Round 1

Questions: 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 i such that 0 ≤ i < n - 1 and swap arr[i] and arr[i + 1]. Each element of the array can be swapped at most once during the whole process.

The strength of an index i is defined as (arr[i] + 1) * (i + 1) using 0-based indexing. Find the maximum possible sum of the strength of all indices after optimal swaps. Mathematically, maximize the following:

Sum = Σ (arr[i] * (i + 1))

Example: Consider n = 4, arr = [2, 1, 4, 3]. It is optimal to swap arr[2] and arr[3] and arr[0] and arr[1]. The final array is [1, 2, 3, 4]. The sum of strengths is 1*1 + 2*2 + 3*3 + 4*4 = 30, which is the maximum possible. Thus, the answer is 30.

Function Description: Complete the function getMaximumSumOfStrengths in the editor below.

getMaximumSumOfStrengths has the following parameter:

• int arr[]: the initial array.

Testcases:

• arr[] = [1, 9, 7, 3, 2] → Output is 66

  • Explanation: It is optimal to swap arr[2] and arr[3]. The final array is arr[] = [1, 9, 3, 7, 2]. The sum of strengths is 1*1 + 2*9 + 3*3 + 4*7 + 5*2 = 66.

• arr[] = [2, 1, 4, 3] → Output is 30

• arr[] = [1, 2, 5] → Output is 20