'''
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).
Elements of input arrays can be modified.
'''
def solution(A):
    # write your code in Python 2.7
    # [ 1 2 3 4 5 6]

    A.sort(reverse=True)
    # partition
    # 1 < P < N-1
    if len(A) == 0:
        return 0
    if len(A) < 3:
        return 1

    prev_d=-1

    for p in xrange(len(A)-1):
        p+=1
        L=sum(A[:p])
        R=sum(A[p:])
        d =abs(L-R)
        print d
        if prev_d == -1:
           prev_d=d
           continue

        if d < prev_d:
           prev_d = d
        else:
           break

    return prev_d

from random import randint
N=100
A=[]
for x in xrange(N):
  A.append(randint(0,N))
print A

print "SOL %i" % solution(A)

