Problem: TapeEquilibrium

P

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 that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.`

`int solution(int A[], int N);`

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;
}``````
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