Codillity Tape equilibrium Solution in Python

The intuitive thing to do is to sum up the array ranges to calculate P of 1..N but that is inefficient.

Observations:
p0 = | (A[0]) - (A[1] + A[2]) |
p1 = | (A[0] + A[1]) - (A[2]) |

If we abstract a bit then we can put the sums into two sets.  The left set can be sum1 and the right set can be sum2.  For each p we essentially add on to set 1 and subtract from set2 the same number.  This saves you from having to sum the array ranges more than you need to.

#solution in Python
import sys
def solution(A):
    sum1 = 0
    sum2 = sum(A)
    diff = sys.maxint
 
    for i in range(0, len(A) -1):
        sum1 += A[i]
        sum2 -= A[i]
     
        _diff = abs(sum1 - sum2)
        if (diff > _diff):
            diff = _diff
    return diff


https://codility.com/demo/results/demo73W5DE-WZP/