Codility[Lesson 3] - Time Complexity(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| = 7P = 2, difference = |4 − 9| = 5P = 3, difference = |6 − 7| = 1P = 4, difference = |10 − 3| = 7

Write a function:

class Solution { public int solution(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].

class Solution {
    public int solution(int[] A) {
        int rightSum = 0;
        int leftSum = 0;
        int result = Integer.MAX_VALUE;

        for(int i=0; i<A.length; i++){
            rightSum += A[i];
        }
        int temp = 0;
        for(int i=1; i<A.length; i++){
            rightSum -= A[i-1];
            leftSum += A[i-1];
            temp = Math.abs(rightSum -leftSum);
            if(temp<result) result = temp;
        }

        return result;
    }
}