TapeEquilibrium

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:

    int solution(vector<int> &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].

#include <limits>

int solution(vector<int> &A) {
 // write your code in C++11 (g++ 4.8.2)
 vector<int> pre_sum;
 int s = 0;
 for (auto i: A) {
        s += i;
        pre_sum.push_back(s);
    }
 
 int min_abs_diff = numeric_limits<int>::max();
 for (size_t i = 1; i < pre_sum.size(); i++) {
 int abs_diff = abs(pre_sum[i - 1] * 2 - pre_sum[pre_sum.size() - 1]);
        min_abs_diff = min(min_abs_diff, abs_diff);
    }
 
 return min_abs_diff;
}