вторник, 24 декабря 2013 г.

Codility. Train. 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 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.

Solution:

def solution(A):
  iMin, iTmp, i1st, i2nd = None, None, 0, sum(A)
  for i in range(len(A) - 1):
    i1st, i2nd = i1st + A[i], i2nd - A[i]
    iTmp = abs(i1st - i2nd)
    if iMin > iTmp or iMin == None:
      iMin = iTmp
  return iMin

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