## wtorek, 5 maja 2015

### Codility solutions - lesson1, exercise: TapeEquilibrium

Content of Codility exercise:

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, A, ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A + A + ... + 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 = 3
A = 1
A = 2
A = 4
A = 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 = 3
A = 1
A = 2
A = 4
A = 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:

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28``` ```public int solution(int[] A) { //initialize variables int minDifference = 0; int divisionNumber = A.length - 1; int leftSum = 0; int rightSum = 0; //calculate table right, left sum and first difference after first division for (int i = 1; i< A.length ; i++){ rightSum = rightSum + A[i]; } leftSum = A; minDifference = Math.abs(leftSum - rightSum); //condition to secure if table A has only two elements if (divisionNumber > 1){ //walking across table A and counting needed variables for (int p = 2; p<=divisionNumber ; p ++){ leftSum = leftSum + A[p-1]; rightSum = rightSum - A[p-1]; int tmpminDifference = Math.abs(leftSum - rightSum); if (minDifference > tmpminDifference){ minDifference = tmpminDifference; } } return minDifference; }else { return minDifference; } } ```