Codility 3.3 TapeEquilibrium

Task description
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].

solution 1: very much time consuming!!! 这个方法有问题，对于A = [-10, -20, -30, -40, 100]这种情况。

A = [-10, -20, -30, -40, 100]

s = sum(A)
i = 0
sum1 = A[0]
sum2 = A[len(A)-1]
while(sum1+sum2 !=s):
    if(sum1<sum2):
        sum1 += A[i+1]
    else:
        sum2 += A[len(A)-i-2]
print(abs(sum1-sum2))

solution 2 :
attention: the sum might be minus

sum1 = A[0]
sum2= sum(A) - A[0]
#consider the element is minus
if len(A) ==2:
    dif = max(A[0]-A[1],A[1]-A[0])
else:
    dif = abs(sum1-sum2)
    for i in range(1,len(A)):    
        sum1 += A[i]
        sum2 -= A[i]
        if dif> abs(sum1-sum2):
            dif = abs(sum1-sum2)