
import java.util.HashMap


/**
 * 100/100
 * @param A
 * @return
 */
fun solution(A: IntArray): Int {
    if (A.size == 0) return -1
    val count: HashMap<Int, Int> = HashMap<Int, Int>()
    for (i in A.indices) {
        if (count.containsKey(A[i])) {
            count[A[i]] = (count[A[i]] ?: 0) + 1
        } else {
            count.put(A[i], 1)
        }
        if ((count[A[i]] ?: 0) > A.size / 2) {
            return i
        }
    }
    return -1
}

/**
 * An array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A.
 *
 * For example, consider array A such that
 *
 *  A[0] = 3    A[1] = 4    A[2] =  3
 *  A[3] = 2    A[4] = 3    A[5] = -1
 *  A[6] = 3    A[7] = 3
 * The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.
 *
 * Write a function
 *
 * class Solution { public int solution(int[] A); }
 *
 * that, given an array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.
 *
 * For example, given array A such that
 *
 *  A[0] = 3    A[1] = 4    A[2] =  3
 *  A[3] = 2    A[4] = 3    A[5] = -1
 *  A[6] = 3    A[7] = 3
 * the function may return 0, 2, 4, 6 or 7, as explained above.
 *
 * Write an efficient algorithm for the following assumptions:
 *
 * N is an integer within the range [0..100,000];
 * each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
 */