FastPrep
Problem · Array

Balanced Sum

● EasyJPMorgan ChaseOA

Given an array of numbers, find the 0-based index of the smallest array element (the pivot) where the sum of all elements to the left and to the right are equal. The array may not be reordered.

Function Description

Complete the function balancedSum in the editor with the following parameter(s):

int arr[n]: an array of integers

Returns

int: the 0-based index of the pivot

Constraints

  • 3≤n≤10^5
  • 1≤arr[i]≤2×10^4, where 0≤i
  • It is guaranteed that a solution always exists.

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Examples
01 · Example 1
arr = [1, 2, 3, 4, 6]
return = 3
The sum of the first three elements is 1 + 2 + 3 = 6. The value of the last element is 6. Using zero-based indexing, arr[3]=4 is the pivot between these two subarrays. Return the index of the pivot, 3.
02 · Example 2
arr = [1, 2, 3, 3]
return = 2
The sum of the first two elements, 1 + 2 = 3. The value of the last element is 3. Using 0-based indexing, arr[2] = 3 is the pivot between these two subarrays. The index of the pivot is 2.
More JPMorgan Chase problems
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public int balancedSum(int[] arr) {
  // write your code here
}
arr[1, 2, 3, 4, 6]
expected3
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