In the presented solution, you are creating and storing two additional arrays, for storing the prefix and suffix sums. For a brute force solution, this is fine but you should provide an optimized version. It's disappointing that such a solution is not given.

There is no need to create additional arrays that cause O(n) space complexity. This can be done in O(1) space by -

1. Storing the left/prefix sums in the result array itself
2. Use an int variable to hold the right/suffix sum and manipulate the result array itself.

Solution -

public class Solution {
    public int[] findDifferenceArray(int[] nums) {
        int n = nums.length;
        int[] differenceArray = new int[n];
        int currLeftSum = 0;
        for(int i = 0; i < n; i++) {
            if(i == 0) {
                differenceArray[i] = 0;
            }
            else {
                currLeftSum += nums[i-1];
                differenceArray[i] = currLeftSum;
            }
        }
        int currRightSum = 0;
        for(int i = n-1; i >= 0; i--) {
            differenceArray[i] = Math.abs(differenceArray[i] - currRightSum);
            currRightSum += nums[i];
        }
        return differenceArray;
    }