Hi,
The problem description is in the comments above the solution below.

The problem belongs to Codility, you can try out it yourself on their platform.

​​​​​​​/**
 * 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. Returns 1 if array A is a permutation and 0 if it is not.
 * 
 */
public class PermCheck {
    public int solution(int[] A) {
        // write your code in Java SE 8
        int[] buck = new int[A.length + 1];
        for (int i = 1; i < buck.length; i++) {
            buck[i] = 0;
        }
        for (int i = 0; i < A.length; i++) {
            if (A[i] < buck.length)
                buck[A[i]]++;
        }
        for (int i = 1; i < buck.length; i++) {
            if (buck[i] != 1) return 0;
        }
        return 1;
    }
}