(Time Complexity) Tape Equilibrium

A non-empty zero-indexed 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 absolute 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 zero-indexed 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.

Assume that: 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].

Complexity: expected worst-case time complexity is O(N); expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).