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(int A[], int N);**

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.h>
int solution(int A[], int N)
{
int sumL = 0;
int sumR = 0;
int minD = INT_MAX;
for(int i=0;i<N;i++)
sumR+=A[i];
for(int i=0;i<N-1;i++)
{
sumL+=A[i];
sumR-=A[i];
int r = abs(sumL-sumR);
minD = minD > r ? r :minD;
}
return minD;
}
```