## 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, A, ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
The difference between the two parts is the value of: |(A + A + ... + 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 = 3
A = 1
A = 2
A = 4
A = 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 = 3
A = 1
A = 2
A = 4
A = 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
```