Minimize the value |(A[0] + … + A[P-1]) – (A[P] + … + A[N-1])|.

## Question

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[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 zero-indexed 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.

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).

## Solution

//A: ("-1000", 1000) expected 2000 //A: (5, 6, 2, 4, 1) expected 4 //A: ("-10", "-5", "-3", "-4", "-5") expected 3 - (int)tapeEquilibrium:(NSMutableArray*) A { int min = 1000000000; //outside bounds int difference; int addup = [A[0]intValue]; int sum = 0; for(int i= 1; i < A.count; i++){ sum += [A[i] integerValue]; } for(int i= 0; i < A.count-1; i++){ difference = abs(addup - sum); if(difference < min || !min){ min = difference; } sum = sum -[A[i+1] intValue]; addup = addup + [A[i+1] intValue]; } return min; }

**Detected time complexity:**

**O(N)**