Codility Perm Check

Problem

A non-empty zero-indexed 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:

int solution(int A[], int N);
that, given a zero-indexed 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.

Assume that:
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].

Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

Solution

public class PermCheck {

	public int solution(int[] A) 
	{
		int invalid = 0, valid =1;
		int n = A.length;
		boolean[] perm = new boolean[n+1];
		int numbers = n;
		
		for(int i=0; i< n; i++)
		{
			if(A[i]> 0 && A[i]<= n)
			{
				if(perm[A[i]] == false) {
					perm[A[i]] = true;
					numbers--;
					if(numbers == 0) return valid;
				}
				else return invalid;
			}
		}
		
		return invalid;
	}

	public static void main(String[] args) {

		int[] A = new int[1000000];//= {0};

		for(int i=1; i<= 1000000; i++)
		{
			A[i-1] =i; 
		}
		int output = new PermCheck().solution(A);
		System.out.println(output);

	}