
import java.util.Arrays


/**
 * 100/100
 * @param A
 * @return
 */
fun solution(A: IntArray): Int {
    val N = A.size
    Arrays.sort(A)
    /**
     * When we sort an array there are two possible options for the largest product
     * 1) The largest (the last) three elements
     * 2) Combination of two smallest and the largest elements
     * Logic of (1): Obvious
     * Logic of (2): A pair of negatives multiplied returns a positive, which in combination with
     * the largest positive element of the array can give the max outcome.
     * Therefore we return the max of options (1) and (2)
     */
    return Math.max(A[0] * A[1] * A[N - 1], A[N - 1] * A[N - 2] * A[N - 3])
}

/**
 * A non-empty array A consisting of N integers is given. The product of triplet (P, Q, R) equates to A[P] * A[Q] * A[R] (0 ≤ P < Q < R < N).
 *
 * For example, array A such that:
 *
 *   A[0] = -3
 *   A[1] = 1
 *   A[2] = 2
 *   A[3] = -2
 *   A[4] = 5
 *   A[5] = 6
 * contains the following example triplets:
 *
 * (0, 1, 2), product is −3 * 1 * 2 = −6
 * (1, 2, 4), product is 1 * 2 * 5 = 10
 * (2, 4, 5), product is 2 * 5 * 6 = 60
 * Your goal is to find the maximal product of any triplet.
 *
 * Write a function:
 *
 * class Solution { public int solution(int[] A); }
 *
 * that, given a non-empty array A, returns the value of the maximal product of any triplet.
 *
 * For example, given array A such that:
 *
 *   A[0] = -3
 *   A[1] = 1
 *   A[2] = 2
 *   A[3] = -2
 *   A[4] = 5
 *   A[5] = 6
 * the function should return 60, as the product of triplet (2, 4, 5) is maximal.
 *
 * Write an efficient algorithm for the following assumptions:
 *
 * N is an integer within the range [3..100,000];
 * each element of array A is an integer within the range [−1,000..1,000].
 */