Lesson3-Tape-Equilibrium.py

# 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:
# 
# def solution(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):
    # write your code in Python 3.6
    left = 0
    minimal = None
    right = sum(A)

    for P in range(0, len(A)-1):
        left += A[P]
        right -= A[P]
        diff = abs(left - right)

        if minimal is None or diff < minimal:
            minimal = diff

    return minimal