/**
 * 100/100
 * @param A
 * @return
 */
fun solution(A: IntArray): Int {
    var left = 0
    var right = A.sum()
    var min = 0
    for (i in 1 until A.size) {
        left += A[i - 1]
        right -= A[i - 1]
        val diff: Int = Math.abs(left - right)
        min = getMin(i, diff, min)
    }
    return min
}

private fun getMin(i: Int, diff: Int, min: Int): Int {
    return if (i == 1 || diff < min) {
        diff
    } else min
}

/**
 * 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:
 *
 * 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].
 */