tape_equilibrium.py

 """
Task: TapeEquilibrium
Goal: Minimize the value |(A[0] + ... + A[P-1]) - (A[P] + ... + A[N-1])|.
Website: https://app.codility.com/programmers/lessons/3-time_complexity/tape_equilibrium/
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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 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:

def solution(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).       
"""


def solution(A):
    differences = []
    length = len(A)
    for index in range(1,length): # Can be split at index 1 through max index
        differences.append(abs(sum(A[:index]) - sum(A[index:])))
    
    return min(differences)