""" A non-empty array A consisting of N integers is given. Array A represents numbers on a tape. Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1]. The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])| In other words, it is the absolute difference between the sum of the first part and the sum of the second part. For example, consider array A such that: A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3 We can split this tape in four places: P = 1, difference = |3 − 10| = 7 P = 2, difference = |4 − 9| = 5 P = 3, difference = |6 − 7| = 1 P = 4, difference = |10 − 3| = 7 Write a function: class Solution { public int solution(int[] A); } that, given a non-empty array A of N integers, returns the minimal difference that can be achieved. For example, given: A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3 the function should return 1, as explained above. Write an efficient algorithm for the following assumptions: N is an integer within the range [2..100,000]; each element of array A is an integer within the range [−1,000..1,000]. """ def solution(A): size = len(A) - 1 if size == 1: return abs(A[0] - A[1]) l = [A[0]] r = [A[-1]] for i in range(size): l.append(l[i] + A[i+1]) r.append(r[i] + A[size-i-1]) res = abs(l[0] - r[size-1]) for i in range(1, size): res = min(res, abs(l[i] - r[size-i-1])) return res res = solution([1, 1, 3]) assert res == 1, "got {}".format(res) res = solution([3, 1, 2, 4, 3]) assert res == 1, "got {}".format(res) res = solution([-2, -3, -4, -1]) assert res == 0, "got {}".format(res) res = solution([1, 8, -10, 8, 7, 9, 11]) assert res == 6, "got {}".format(res) res = solution([-1, 1, 0, 1, -1, 0, -1, 1, -1, 0, 1, 0, 1, -1, 0]) assert res == 0, "got {}".format(res) res = solution([14, 9]) assert res == 5, "got {}".format(res)