"""
My solution:
Time Complexity:
The time complexity is O(n),
where n is the length of the input list nums.
The function iterates through the list twice: once to calculate the total sum (total),
and once to calculate the difference array.
Space Complexity:
The space complexity is O(n).
The differenceArray list is of the same length as nums,
and additional variables (total, left, right, diff_total, el) use constant space.
"""
class Solution:
    def findDifferenceArray(self, nums):
        n = len(nums)
        differenceArray = [0] * n

        total = 0
        for el in nums:
            total += el

        left = 0
        for idx, el in enumerate(nums):
            left += el
            right = total - left
            diff_total = (left - el) - right
            differenceArray[idx] = diff_total * -1 if diff_total < 0 else diff_total

        return differenceArray