/**
 * 100/100
 * @param A
 * @return
 */
fun solution(A: IntArray): Int {
    val uniqueCheck = IntArray(A.size)
    for (i in A.indices) {
        val current = A[i] - 1
        if (A[i] > A.size) return 0
        uniqueCheck[current]++
        if (uniqueCheck[current] > 1) return 0
    }
    return 1
}

/**
 * A non-empty array A consisting of N integers is given.
 *
 * A permutation is a sequence containing each element from 1 to N once, and only once.
 *
 * For example, array A such that:
 *
 *     A[0] = 4
 *     A[1] = 1
 *     A[2] = 3
 *     A[3] = 2
 * is a permutation, but array A such that:
 *
 *     A[0] = 4
 *     A[1] = 1
 *     A[2] = 3
 * is not a permutation, because value 2 is missing.
 *
 * The goal is to check whether array A is a permutation.
 *
 * Write a function:
 *
 * class Solution { public int solution(int[] A); }
 *
 * that, given an array A, returns 1 if array A is a permutation and 0 if it is not.
 *
 * For example, given array A such that:
 *
 *     A[0] = 4
 *     A[1] = 1
 *     A[2] = 3
 *     A[3] = 2
 * the function should return 1.
 *
 * Given array A such that:
 *
 *     A[0] = 4
 *     A[1] = 1
 *     A[2] = 3
 * the function should return 0.
 *
 * Write an efficient algorithm for the following assumptions:
 *
 * N is an integer within the range [1..100,000];
 * each element of array A is an integer within the range [1..1,000,000,000].
 */